Chapter 21: Trigonometrical Identities

Exercise 21(A)

Question 1(a)

sin⁴ θ - cos⁴ θ is equal to :

sin² θ - 1
1 + cos² θ
2 sin² θ - 1
1 - cos² θ

Answer

Solving,

⇒ sin⁴ θ - cos⁴ θ

⇒ (sin² θ + cos² θ)(sin² θ - cos² θ)

Substituting, sin² θ + cos² θ = 1, we get :

⇒ 1 × (sin² θ - cos² θ)

⇒ sin² θ - cos² θ

⇒ sin² θ - (1 - sin² θ)

⇒ sin² θ + sin² θ - 1

⇒ 2 sin² θ - 1.

Hence, Option 3 is the correct option.

Question 1(b)

(1 + tan θ)² + (1 - tan θ)² is equal to :

2 cosec² θ
2 sec² θ
cosec² θ
sec² θ

Answer

Solving,

⇒ (1 + tan θ)² + (1 - tan θ)²

⇒ 1 + tan² θ + 2 tan θ + 1 + tan² θ - 2 tan θ

⇒ 2 + 2 tan² θ

⇒ 2(1 + tan² θ)

Substituting, 1 + tan² θ = sec² θ, we get :

⇒ 2 sec² θ.

Hence, Option 2 is the correct option.

Question 1(c)

cot⁴ θ + cot² θ is equal to :

2 cot² θ. cosec² θ
tan² θ + tan⁴ θ
tan² θ.cosec² θ
cosec⁴ θ - cosec² θ

Answer

Solving,

⇒ cot⁴ θ + cot² θ

⇒ cot² θ(cot² θ + 1)

⇒ cot² θ.cosec² θ

Substituting, cot² θ = cosec² θ - 1, we get :

⇒ (cosec² θ - 1).cosec² θ

⇒ cosec⁴ θ - cosec² θ.

Hence, Option 4 is the correct option.

Question 1(d)

$\dfrac{1}{\text{1 - sin A}}$ is equal to :

1 + sin A
sec² A (1 + sin A)
sec² A
sec² A + sin A

Answer

Solving,

⇒ $\dfrac{1}{\text{1 - sin A}}$

Multiplying numerator and denominator by (1 + sin A), we get :

$\Rightarrow \dfrac{1}{\text{1 - sin A}} \times \dfrac{\text{1 + sin A}}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A}}{1 - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{1 + sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 A} + \dfrac{\text{sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{cos}^2 A}. \text{ sin A} \\[1em] \Rightarrow \text{sec}^2 A + \text{sec}^2 A.\text{ sin A} \\[1em] \Rightarrow \text{sec}^2 A\text{(1 + sin A)}.$

Hence, Option 2 is the correct option.

Question 1(e)

$\dfrac{\text{tan}^2 A}{\text{(sec A + 1)}^2}$ is equal to :

$\dfrac{\text{1 + cos A}}{\text{1 - cos A}}$
$\dfrac{\text{1 - cos A}}{\text{1 + cos A}}$
$\dfrac{1}{\text{1 + cos A}}$
$\dfrac{1}{\text{1 - cos A}}$

Answer

Solving,

$\Rightarrow \dfrac{\text{tan}^2 A}{\text{(sec A + 1)}^2} \\[1em] \Rightarrow \dfrac{\text{sec}^2 A - 1}{\text{(sec A + 1)}^2} \\[1em] \Rightarrow \dfrac{\text{(sec A + 1)(sec A - 1)}}{\text{(sec A + 1)}^2} \\[1em] \Rightarrow \dfrac{\text{sec A - 1}}{\text{sec A + 1}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos A}} - 1}{\dfrac{1}{\text{cos A}} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - cos A}}{\text{cos A}}}{\dfrac{\text{1 + cos A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A}}{\text{1 + cos A}}.$

Hence, Option 2 is the correct option.

Question 2

Prove the following identities :

$\dfrac{\text{sec A - 1}}{\text{sec A + 1}} = \dfrac{\text{1 - cos A}}{\text{1 + cos A}}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{sec A - 1}}{\text{sec A + 1}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos A}} - 1}{\dfrac{1}{\text{cos A}} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - cos A}}{\text{cos A}}}{\dfrac{\text{1 + cos A}}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{(1 - cos A) × cos A}}{\text{(1 + cos A) × cos A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A}}{\text{1 + cos A}}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sec A - 1}}{\text{sec A + 1}} = \dfrac{\text{1 - cos A}}{\text{1 + cos A}}$ .

Question 3

Prove the following identities :

$\dfrac{1}{\text{tan A + cot A}} = \text{cos A sin A}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1}{\text{tan A + cot A}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}} \\[1em] \Rightarrow \dfrac{\text{sin A cos A}}{\text{sin}^2 A + \text{cos}^2 A}$

By formula,

sin² A + cos² A = 1

$\Rightarrow \text{sin A cos A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{1}{\text{tan A + cot A}} = \text{cos A sin A}$ .

Question 4

Prove the following identities :

tan A - cot A = $\dfrac{\text{1 - 2 cos}^2 A}{\text{sin A cos A}}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \text{tan A - cot A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} - \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{cos}^2 A}{\text{cos A sin A}}$

By formula,

sin² A = 1 - cos² A

$\Rightarrow \dfrac{1 - \text{cos}^2 A - \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{1 - \text{2 cos}^2 A}{\text{sin A cos A}}.$

Since, L.H.S. = R.H.S.

Hence, proved that tan A - cot A = $\dfrac{\text{1 - 2 cos}^2 A}{\text{sin A cos A}}$

Question 5

Prove the following identities :

cosec⁴ A - cosec² A = cot⁴ A + cot² A

Answer

Solving L.H.S. of the equation :

⇒ cosec⁴ A - cosec² A

⇒ cosec² A(cosec² A - 1)

By formula,

cosec² A = 1 + cot² A

⇒ (1 + cot² A)(1 + cot² A - 1)

⇒ (1 + cot² A)cot² A

⇒ cot² A + cot⁴ A.

Since, L.H.S. = R.H.S.

Hence, proved that cosec⁴ A - cosec² A = cot⁴ A + cot² A.

Question 6

Prove the following identities :

sec A(1 - sin A)(sec A + tan A) = 1

Answer

Solving L.H.S. of the equation :

$\Rightarrow \text{sec A(1 - sin A)(sec A + tan A)} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} \times (\text{1 - sin A}) \times \Big(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big) \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} \times (\text{1 - sin A}) \times \Big(\dfrac{\text{1 + sin A}}{\text{cos A}}\Big) \\[1em] \Rightarrow \dfrac{1 - \text{sin}^2 A}{\text{cos}^2A}$

By formula,

cos² A = 1 - sin² A

$\Rightarrow \dfrac{1 - \text{sin}^2 A}{1 - \text{sin}^2A} \\[1em] \Rightarrow 1.$

Since, L.H.S. = R.H.S.

Hence, proved that sec A(1 - sin A)(sec A + tan A) = 1.

Question 7

Prove the following identities :

sec² A + cosec² A = sec² A . cosec² A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \text{sec}^2 A + \text{cosec}^2 A \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos}^2 A. \text{sin}^2 A}$

As, sin² A + cos² A = 1

$\Rightarrow \dfrac{1}{\text{cos}^2 A. \text{sin}^2 A} \\[1em] \Rightarrow \text{sec}^2 A. \text{cosec}^2 A$

Since, L.H.S. = R.H.S.

Hence, proved that sec² A + cosec² A = sec² A . cosec² A.

Question 8

Prove the following identities :

$\dfrac{\text{(1 + tan}^2 A)\text{cot A}}{\text{cosec}^2 A} = \text{tan A}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\Big(1 + \dfrac{\text{sin}^2 A}{\text{cos}^2 A}\Big) \times \dfrac{\text{cos A}}{\text{sin A}}}{\dfrac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \Big(\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}\Big) \times \dfrac{\text{cos A}}{\text{sin A}} \times \text{sin}^2 A$

By formula,

cos² A + sin² A = 1

$\Rightarrow \dfrac{1}{\text{cos}^2 A} \times \text{cos A. sin A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{tan A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{(1 + tan}^2 A)\text{cot A}}{\text{cosec}^2 A} = \text{tan A}$ .

Question 9

Prove the following identities :

tan² A - sin² A = tan² A. sin² A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \text{tan}^2 A - \text{sin}^2 A \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \text{sin}^2 A \\[1em] \Rightarrow \text{sin}^2 A \times \Big(\dfrac{1}{\text{cos}^2 A} - 1\Big) \\[1em] \Rightarrow \text{sin}^2 A \times \Big(\dfrac{1 - \text{cos}^2 A}{\text{cos}^2 A}\Big)$

By formula,

1 - cos²A = sin² A.

$\Rightarrow \text{sin}^2 A \times \dfrac{\text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \text{sin}^2 A. \text{tan}^2 A$

Since, L.H.S. = R.H.S.

Hence, proved that tan² A - sin² A = tan² A. sin² A.

Question 10

Prove the following identities :

(cosec A + sin A)(cosec A - sin A) = cot² A + cos² A

Answer

By formula,

cosec² A = 1 + cot² A

sin² A = 1 - cos² A

Solving L.H.S. of the equation

⇒ (cosec A + sin A)(cosec A - sin A)

⇒ cosec² A - sin² A

⇒ 1 + cot² A - (1 - cos² A)

⇒ 1 - 1 + cot² A + cos² A

⇒ cot² A + cos² A.

Hence, proved that (cosec A + sin A)(cosec A - sin A) = cot² A + cos² A.

Question 11

Prove the following identities :

(cos A + sin A)² + (cos A - sin A)² = 2

Answer

Solving L.H.S. of the equation :

⇒ (cos A + sin A)² + (cos A - sin A)²

⇒ cos² A + sin² A + 2 cos A sin A + cos² A + sin² A - 2 cos A sin A

⇒ 2(sin² A + cos² A) + 2cos A sin A - 2cos A sin A

As, sin² A + cos² A = 1

⇒ 2 × 1

⇒ 2.

Since, L.H.S. = R.H.S.

Hence, proved that (cos A + sin A)² + (cos A - sin A)² = 2.

Question 12

Prove the following identities :

(cosec A - sin A)(sec A - cos A)(tan A + cot A) = 1

Answer

Solving L.H.S. of the equation :

$\Rightarrow \text{(cosec A - sin A)(sec A - cos A)(tan A + cot A)} \\[1em] \Rightarrow \Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big) \times \Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big) \times \Big(\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}\Big) \\[1em] \Rightarrow \Big(\dfrac{1 - \text{sin}^2 A}{\text{sin A}}\Big) \times \Big(\dfrac{1 - \text{cos}^2 A}{\text{cos A}}\Big) \times \Big(\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A. sin A}}\Big)$

By formula,

1 - sin² A = cos² A, 1 - cos² A = sin² A and sin² A + cos² A = 1.

$\Rightarrow \Big(\dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}} \times \dfrac{1}{\text{cos A. sin A}}\Big) \\[1em] \Rightarrow \dfrac{\text{cos}^2 A. \text{sin}^2 A}{\text{cos}^2 A. \text{sin}^2 A} \\[1em] \Rightarrow 1.$

Since, L.H.S. = R.H.S.

Hence, proved that (cosec A - sin A)(sec A - cos A)(tan A + cot A) = 1.

Question 13

Prove the following identities :

$\dfrac{1}{\text{sec A + tan A}}$ = sec A - tan A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1}{\text{sec A + tan A}}$

Multiplying numerator and denominator by (sec A - tan A), we get :

$\Rightarrow \dfrac{1}{\text{sec A + tan A}} \times \dfrac{\text{sec A - tan A}}{\text{sec A - tan A}} \\[1em] \Rightarrow \dfrac{\text{sec A - tan A}}{\text{sec}^2 A - \text{tan}^2 A}$

By formula,

sec² A - tan² A = 1

$\Rightarrow \dfrac{\text{sec A - tan A}}{1} \\[1em] \Rightarrow \text{sec A - tan A}.$

Since, L.H.S. = R.H.S.

Hence proved that $\dfrac{1}{\text{sec A + tan A}}$ = sec A - tan A.

Question 14

Prove the following identities :

(sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

Answer

By formula,

sin² A + cos² A = 1

sec² A = 1 + tan² A

cosec² A = 1 + cot² A

Solving L.H.S. of the equation :

⇒ (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

⇒ sin² A + cosec² A + 2 sin A. cosec A + cos² A + sec² A + 2 cos A. sec A

⇒ sin² A + 1 + cot² A + 2 × sin A × $\dfrac{1}{\text{sin A}}$ + cos² A + 1 + tan² A + 2 × cos A × $\dfrac{1}{\text{cos A}}$

⇒ sin² A + cos² A + 1 + cot² A + 2 + 1 + tan² A + 2

⇒ 1 + 1 + 2 + 1 + 2 + cot² A + tan² A

⇒ 7 + tan² A + cot² A.

Since, L.H.S. = R.H.S.

Hence, proved that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A.

Question 15

Prove the following identities :

sec² A . cosec² A = tan² A + cot² A + 2

Answer

Solving L.H.S. of the equation :

⇒ sec² A . cosec² A

⇒ $\dfrac{1}{\text{cos}^2 A. \text{ sin}^2 A}$ .

Solving R.H.S. of the equation :

$\Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} + \dfrac{\text{cos}^2 A}{\text{sin}^2 A} + 2 \\[1em] \Rightarrow \dfrac{\text{sin}^4 A + \text{cos}^4 A + \text{2 sin}^2 A \text{ cos}^2 A}{\text{cos}^2 A \text{ sin}^2 A} \\[1em] \Rightarrow \dfrac{(\text{sin}^2 A + \text{cos}^2 A)^2}{\text{cos}^2 A \text{ sin}^2 A}$

By formula,

sin² A + cos² A = 1.

$\Rightarrow \dfrac{(1)^2}{\text{cos}^2 A \text{ sin}^2 A} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 A \text{ sin}^2 A}.$

Since, L.H.S. = R.H.S. = $\dfrac{1}{\text{cos}^2 A \text{ sin}^2 A}.$

Hence, proved that sec² A . cosec² A = tan² A + cot² A + 2.

Question 16

Prove the following identities :

$\dfrac{1}{\text{1 + cos A}} + \dfrac{1}{\text{1 - cos A}}$ = 2 cosec² A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1}{\text{1 + cos A}} + \dfrac{1}{\text{1 - cos A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A + 1 + cos A}}{\text{(1 + cos A)(1 - cos A)}} \\[1em] \Rightarrow \dfrac{2}{1 - \text{cos}^2 A}$

By formula,

1 - cos² A = sin² A

$\Rightarrow \dfrac{2}{\text{sin}^2 A} \\[1em] \Rightarrow 2\text{ cosec}^2 A.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{1}{\text{1 + cos A}} + \dfrac{1}{\text{1 - cos A}}$ = 2 cosec² A.

Question 17

Prove the following identities :

$\dfrac{\text{cosec A}}{\text{cosec A - 1}} + \dfrac{\text{cosec A}}{\text{cosec A + 1}}$ = 2 sec² A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{cosec A}}{\text{cosec A - 1}} + \dfrac{\text{cosec A}}{\text{cosec A + 1}} \\[1em] \Rightarrow \dfrac{\text{cosec A(cosec A + 1) + cosec A(cosec A - 1)}}{\text{(cosec A - 1)(cosec A + 1)}} \\[1em] \Rightarrow \dfrac{\text{cosec}^2 A + \text{cosec A + cosec}^2 A - \text{cosec A}}{\text{cosec}^2 A - 1} \\[1em] \Rightarrow \dfrac{\text{2 cosec}^2 A}{\text{cot}^2 A} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{\text{sin}^2 A}}{\dfrac{\text{cos}^2 A}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{\text{sin}^2 A} \times \text{sin}^2 A}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{2}{\text{cos}^2 A} \\[1em] \Rightarrow 2\text{sec}^2 A.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cosec A}}{\text{cosec A - 1}} + \dfrac{\text{cosec A}}{\text{cosec A + 1}}$ = 2 sec² A.

Question 18

Prove the following identities :

$\dfrac{\text{1 + cos A}}{\text{1 - cos A}} = \dfrac{\text{tan}^2 A}{\text{(sec A - 1)}^2}$

Answer

Solving R.H.S. of the equation :

$\Rightarrow \dfrac{\text{tan}^2 A}{\text{(sec A - 1)}^2} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin}^2 A}{\text{cos}^2 A}}{\Big(\dfrac{1}{\text{cos A}} - 1\Big)^2} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin}^2 A}{\text{cos}^2 A}}{\Big(\dfrac{\text{1 - cos A}}{\text{cos} A}\Big)^2} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin}^2 A}{\text{cos}^2 A} \times \text{cos}^2 A}{\text{(1 - cos A)}^2} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{(1 - cos A)}^2}.$

By formula,

sin² A = 1 - cos² A

$\Rightarrow \dfrac{\text{1 - cos}^2 A}{(1 - \text{ cos A})^2} \\[1em] \Rightarrow \dfrac{\text{(1 - cos A)(1 + cos A)}}{\text{(1 - cos A)}^2} \\[1em] \Rightarrow \dfrac{\text{(1 + cos A)}}{\text{(1 - cos A)}}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{1 + cos A}}{\text{1 - cos A}} = \dfrac{\text{tan}^2 A}{\text{(sec A - 1)}^2}$ .

Question 19

Prove the following identities :

$\dfrac{\text{1 + sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{1 + sin A}}$ = 2 sec A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{(1 + sin A)}^2 + \text{cos}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{1 + 2 sin A + sin}^2 A + \text{cos}^2 A}{\text{cos A(1 + sin A)}}$

By formula,

sin² A + cos² A = 1.

$\Rightarrow \dfrac{\text{1 + 2 sin A + 1}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{2 + 2 sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{2(1 + sin A)}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{2}{\text{cos A}} \\[1em] \Rightarrow 2\text{ sec A}$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{1 + sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{1 + sin A}}$ = 2 sec A.

Question 20

Prove the following identities :

$\dfrac{\text{1 - sin A}}{\text{1 + sin A}}$ = (sec A - tan A)²

Answer

Solving R.H.S. of the equation :

$\Rightarrow \text{(sec A - tan A)}^2 \\[1em] \Rightarrow \Big(\dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{1 - sin A}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)}^2}{\text{cos}^2 A}$

By formula,

cos² A = 1 - sin² A

$\Rightarrow \dfrac{\text{(1 - sin A)}^2}{\text{1 - sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)}^2}{\text{(1 - sin A)(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{1 - sin A}}{\text{1 + sin A}}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{1 - sin A}}{\text{1 + sin A}}$ = (sec A - tan A)².

Question 21

Prove the following identities :

$\dfrac{\text{cosec A - 1}}{\text{cosec A + 1}} = \Big(\dfrac{\text{cos A}}{\text{1 + sin A}}\Big)^2$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\dfrac{1}{\text{sin A}} - 1}{\dfrac{1}{\text{sin A}} + 1} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - sin A}}{\text{sin A}}}{\dfrac{\text{1 + sin A}}{\text{sin A}}} \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)} \times \text{ sin A}}{\text{(1 + sin A)} \times \text{ sin A}} \\[1em] \Rightarrow \dfrac{\text{1 - sin A}}{\text{1 + sin A}}.$

Solving R.H.S. of the equation :

$\Rightarrow \Big(\dfrac{\text{cos A}}{\text{1 + sin A}}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{(1 + sin A)}^2} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 A}{\text{(1 + sin A)}^2} \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)(1 + sin A)}}{\text{(1 + sin A)}^2} \\[1em] \Rightarrow \dfrac{\text{(1 - sin A)}}{\text{(1 + sin A)}}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cosec A - 1}}{\text{cosec A + 1}} = \Big(\dfrac{\text{cos A}}{\text{1 + sin A}}\Big)^2$ .

Question 22

Prove the following identities :

tan² A - tan² B = $\dfrac{\text{sin}^2 A - \text{sin}^2 B}{\text{cos}^2 A. \text{cos}^2 B}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \dfrac{\text{sin}^2 B}{\text{cos}^2 B} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A. \text{ cos}^2 B - \text{sin}^2 B. \text{ cos}^2 A}{\text{cos}^2 A. \text{ cos}^2 B} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A(1 - \text{ sin}^2 B) - \text{sin}^2 B(1 - \text{sin}^2 A)}{\text{cos}^2 A. \text{ cos}^2 B} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{sin}^2 A. \text{ sin}^2 B - \text{sin}^2 B + \text{ sin}^2 A. \text{ sin}^2 B}{\text{cos}^2 A. \text{ cos}^2 B} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{ sin}^2 B}{\text{cos}^2 A. \text{ cos}^2 B}$

Since, L.H.S. = R.H.S.

Hence, proved that tan² A - tan² B = $\dfrac{\text{sin}^2 A - \text{sin}^2 B}{\text{cos}^2 A. \text{cos}^2 B}$ .

Question 23

Prove the following identities :

$\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{ cos θ}}$ = tan θ

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{sin θ(1 - 2 sin}^2 θ)}{\text{cos θ(2 cos}^2 θ - 1)}$

By formula,

sin² θ = 1 - cos² θ

$\Rightarrow \dfrac{\text{sin θ[1 - 2(1 - cos}^2 θ)]}{\text{cos θ(2 cos}^2 θ - 1)} \\[1em] \Rightarrow \text{tan θ} \times \dfrac{1 - 2 + \text{ 2 cos}^2 θ}{\text{(2 cos}^2 θ - 1)} \\[1em] \Rightarrow \text{tan θ} \times \dfrac{\text{2 cos}^2 θ - 1}{\text{2 cos}^2 θ - 1} \\[1em] \Rightarrow \text{tan θ}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin θ - 2 sin}^3 θ}{\text{2 cos}^3 θ - \text{ cos θ}}$ = tan θ.

Question 24

Prove the following identities :

$\dfrac{\text{cos A}}{\text{1 - sin A}}$ = sec A + tan A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{cos A}}{\text{1 - sin A}}$

Multiplying numerator and denominator by (1 + sin A)

$\Rightarrow \dfrac{\text{cos A(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos A(1 + sin A)}}{\text{1 - sin}^2 A}$

By formula,

cos² A = 1 - sin² A

$\Rightarrow \dfrac{\text{cos A(1 + sin A)}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{1 + sin A}}{\text{\text{cos A}}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + tan A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cos A}}{\text{1 - sin A}}$ = sec A + tan A.

Question 25

Prove the following identities :

$\dfrac{\text{sin A tan A}}{\text{1 - cos A}}$ = 1 + sec A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{sin A} \times \dfrac{\text{sin A}}{\text{cos A}}}{\text{1 - cos A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{cos A(1 - cos A)}}$

By formula,

sin² A = 1 - cos² A

$\Rightarrow \dfrac{1 - \text{cos}^2 A}{\text{cos A(1 - cos A)}} \\[1em] \Rightarrow \dfrac{\text{(1 - cos A)(1 + cos A)}}{\text{cos A(1 - cos A)}} \\[1em] \Rightarrow \dfrac{\text{1 + cos A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{cos A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + 1}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin A tan A}}{\text{1 - cos A}}$ = 1 + sec A.

Question 26

Prove the following identities :

(1 + cot A - cosec A)(1 + tan A + sec A) = 2

Answer

Solving L.H.S. of the equation :

$\Rightarrow \Big(1 + \dfrac{\text{cos A}}{\text{sin A}} - \dfrac{1}{\text{sin A}}\Big)\Big(1 + \dfrac{\text{sin A}}{\text{cos A}}+ \dfrac{1}{\text{cos A}}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{sin A + cos A - 1}}{\text{sin A}}\Big)\Big(\dfrac{\text{cos A + sin A + 1}}{\text{cos A}}\Big) \\[1em] \Rightarrow \dfrac{\text{(sin A + cos A - 1)(sin A + cos A + 1)}}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{sin A cos A + sin A + cos A sin A + cos A} + \text{ cos}^2 A - \text{sin A - cos A - 1}}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + \text{2 cos A sin A - 1}}{\text{sin A cos A}}.$

By formula,

sin² A + cos² A = 1.

$\Rightarrow \dfrac{1 + \text{2 cos A sin A - 1}}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{2 cos A sin A}}{\text{cos A sin A}} \\[1em] \Rightarrow 2.$

Since, L.H.S. = R.H.S.

Hence, proved that (1 + cot A - cosec A)(1 + tan A + sec A) = 2.

Question 27

Prove the following identities :

$\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}}$ = sec A + tan A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}}$

Multiplying numerator and denominator by $\sqrt{1 + \text{sin A}}$

$\Rightarrow \sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} \times \sqrt{\dfrac{\text{1 + sin A}}{\text{1 + sin A}}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 + sin A)(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 + sin A)(1 + sin A)}}{\text{1 - sin}^2 \text{A}}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 + sin A)}^2}{\text{cos}^2 A}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + tan A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}}$ = sec A + tan A.

Question 28

Prove the following identities :

$\sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A - cot A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A - cot A

Multiplying numerator and denominator by $\sqrt{1 - \text{cos A}}$

$\Rightarrow \sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} \times \sqrt{\dfrac{\text{1 - cos A}}{\text{1 - cos A}}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 - cos A)(1 - cos A)}}{\text{(1 + cos A)(1 - cos A)}}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 - cos A)}^2}{\text{(1 - cos}^2 A)}}$

By formula,

1 - cos² A = sin² A

$\Rightarrow \sqrt{\dfrac{\text{(1 - cos A)}^2}{\text{sin}^2 A}} \\[1em] \Rightarrow \dfrac{\text{1 - cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}} - \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A - cot A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}}$ = cosec A - cot A.

Question 29

Prove the following identities :

1 - $\dfrac{\text{cos}^2 A}{\text{1 + sin A}}$ = sin A

Answer

Solving L.H.S. of the equation :

$\Rightarrow 1 - \dfrac{\text{cos}^2 A}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A - cos}^2 A}{\text{1 + sin A}}$

By formula,

cos² A = 1 - sin² A

$\Rightarrow \dfrac{\text{1 + sin A - (1 - sin}^2 A)}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A - 1 + sin}^2 A}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{sin A + sin}^2 A}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{sin A(1 + sin A)}}{\text{1 + sin A}} \\[1em] \Rightarrow \text{sin A}.$

Since, L.H.S. = R.H.S.

Hence, proved that 1 - $\dfrac{\text{cos}^2 A}{\text{1 + sin A}}$ = sin A.

Question 30

Prove the following identities :

$\dfrac{1}{\text{sin A + cos A}} + \dfrac{1}{\text{sin A - cos A}} = \dfrac{\text{2 sin A}}{\text{1 - 2 cos}^2 A}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{sin A - cos A + sin A + cos A}}{\text{(sin A + cos A)(sin A - cos A)}} \\[1em] \Rightarrow \dfrac{\text{2 sin A}}{\text{sin}^2 A - \text{cos}^2 A}$

By formula,

sin² A = 1 - cos² A

$\Rightarrow \dfrac{\text{2 sin A}}{\text{1 - cos}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 sin A}}{\text{1 - 2 cos}^2 A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{1}{\text{sin A + cos A}} + \dfrac{1}{\text{sin A - cos A}} = \dfrac{\text{2 sin A}}{\text{1 - 2 cos}^2 A}$ .

Question 31

Prove the following identities :

$\dfrac{\text{sin A + \text{cos A}}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{2 sin}^2 A - 1}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{(sin A + cos A)}^2 + \text{(sin A - cos A)}^2}{\text{(sin A - cos A)(sin A + cos A)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 + \text{2 sin A cos A} + \text{sin}^2 A + \text{cos}^2 A - \text{2 sin A cos A}}{\text{sin}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 (sin}^2 A + \text{ cos}^2 A)}{\text{sin}^2 A - \text{cos}^2 A}$

By formula,

sin² A + cos² A = 1

cos² A = 1 - sin² A

$\Rightarrow \dfrac{2}{\text{sin}^2 A - (1 - \text{sin}^2 A)} \\[1em] \Rightarrow \dfrac{2}{\text{sin}^2 A - 1 + \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{2}{\text{2 sin}^2 A - 1}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin A + \text{cos A}}}{\text{sin A - cos A}} + \dfrac{\text{sin A - cos A}}{\text{sin A + cos A}} = \dfrac{2}{\text{2 sin}^2 A - 1}$

Question 32

Prove the following identities :

$\dfrac{\text{1 + sin A}}{\text{cosec A - cot A}} - \dfrac{\text{1 - sin A}}{\text{cosec A + cot A}}$ = 2(1 + cot A)

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{1 + sin A}}{\text{cosec A - cot A}} - \dfrac{\text{1 - sin A}}{\text{cosec A + cot A}} \\[1em] \Rightarrow \dfrac{\text{(1 + sin A)(cosec A + cot A)} - \text{(1 - sin A)(cosec A - cot A)}}{\text{cosec}^2 A - \text{cot}^2 A}$

By formula,

cosec² A - cot² A = 1

⇒ (1 + sin A)(cosec A + cot A) - (1 - sin A)(cosec A - cot A)

⇒ cosec A + cot A + sin A cosec A + sin A cot A - (cosec A - cot A - sin A cosec A + sin A cot A)

⇒ cosec A - cosec A + cot A + cot A + sin A cosec A + sin A cosec A + sin A cot A - sin A cot A

⇒ 2 cot A + 2 sin A cosec A

⇒ 2 cot A + $2 \text{ sin A} \times \dfrac{1}{\text{sin A}}$

⇒ 2 cot A + 2

⇒ 2(cot A + 1).

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{1 + sin A}}{\text{cosec A - cot A}} - \dfrac{\text{1 - sin A}}{\text{cosec A + cot A}}$ = 2(1 + cot A).

Question 33

Prove the following identities :

$\dfrac{\text{cos θ cot θ}}{\text{1 + sin θ}}$ = cosec θ - 1

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{cos θ} \times \dfrac{\text{cos θ}}{\text{sin θ}}}{\text{(1 + sin θ)}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ}{\text{sin θ(1 + sin θ)}}$

By formula,

cos² θ = 1 - sin² θ

$\Rightarrow \dfrac{\text{1 - sin}^2 θ}{\text{sin θ(1 + sin θ)}} \\[1em] \Rightarrow \dfrac{\text{(1 - sin θ)(1 + sin θ)}}{\text{sin θ(1 + sin θ)}} \\[1em] \Rightarrow \dfrac{\text{1 - sin θ}}{\text{sin θ}} \\[1em] \Rightarrow \dfrac{1}{\text{sin θ}} - \dfrac{\text{sin θ}}{\text{sin θ}} \\[1em] \Rightarrow \text{cosec θ - 1}$

Since, L.H.S. = R.H.S

Hence, proved that $\dfrac{\text{cos θ cot θ}}{\text{1 + sin θ}}$ = cosec θ - 1.

Exercise 21(B)

Question 1(a)

(1 + cot² A) + (1 + tan² A) is equal to :

$\dfrac{1}{\text{sin}^2 A - \text{cos}^2 A}$
sec² A - cosec² A
sin² A - sin⁴ A
sec² A.cosec² A

Answer

Solving,

⇒ (1 + cot² A) + (1 + tan² A)

⇒ cosec² A + sec² A

⇒ $\dfrac{1}{\text{sin}^2 A} + \dfrac{1}{\text{cos}^2 A}$

⇒ $\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{sin}^2 A. \text{cos}^2 A}$

Substituting, sin² A + cos² A = 1, we get :

⇒ $\dfrac{1}{\text{sin}^2 A. \text{cos}^2 A}$

⇒ cosec² A. sec² A

Hence, Option 4 is the correct option.

Question 1(b)

If a = tan θ and b = sec θ, the relation between a and b is :

a × b = 1
a² - b² = 1
b² - a² = 1
a² + b² = 1

Answer

Substituting value of a and b in L.H.S. of option 3, we get :

⇒ b² - a²

⇒ sec² A - tan² A

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, Option 3 is the correct option.

Question 1(c)

$\dfrac{1}{\text{cot θ + tan θ}}$ is equal to :

sin θ + cos θ
sin θ. cos θ
$\dfrac{1}{\text{sin θ. cos θ}}$
$\dfrac{1}{\text{sin θ + cos θ}}$

Answer

Solving,

$\Rightarrow \dfrac{1}{\text{cot θ + tan θ}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{cos θ}}{\text{sin θ}} + \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\text{cos}^2 θ + \text{sin}^2 θ}{\text{cos θ. sin θ}}} \\[1em] \Rightarrow \dfrac{\text{cos θ. sin θ}}{\text{cos}^2 θ + \text{sin}^2 θ}$

Substituting, sin² θ + cos² θ = 1, we get :

⇒ sin θ. cos θ

Hence, Option 2 is the correct option.

Question 1(d)

(sec θ - cos θ)² - (sec θ + cos θ)² is equal to :

4
2
-2
-4

Answer

Solving,

⇒ (sec θ - cos θ)² - (sec θ + cos θ)²

⇒ (sec θ - cos θ + sec θ + cos θ)[(sec θ - cos θ) - (sec θ + cos θ)] [∵ a² - b² = (a + b)(a - b)]

⇒ (sec θ - cos θ + sec θ + cos θ)(sec θ - sec θ - cos θ - cos θ)

⇒ 2 sec θ.(-2 cos θ)

⇒ -4.sec θ.cos θ

⇒ $-4 \times \dfrac{1}{\text{cos θ}} \times \text{cos θ}$

⇒ -4.

Hence, Option 4 is the correct option.

Question 1(e)

(cot A - cot B)² + (1 + cot A cot B)² is equal to :

sec² A - cos² A
sec² A - cosec² A
(1 + tan A)² - (1 - cot A)²
cosec² A . cosec² B

Answer

Solving,

⇒ (cot A - cot B)² + (1 + cot A cot B)²

⇒ cot² A + cot² B - 2 cot A cot B + 1 + cot² A . cot² B + 2 cot A cot B

⇒ cot² A + cot² A . cot² B + 1 + cot² B

⇒ cot² A(1 + cot² B) + (1 + cot² B)

⇒ (1 + cot² B)(1 + cot² A)

⇒ $\Big(1 + \dfrac{\text{cos}^2 B}{\text{sin}^2 B}\Big)\Big(1 + \dfrac{\text{cos}^2 A}{\text{sin}^2 A}\Big)$

⇒ $\Big(\dfrac{\text{sin}^2 B + \text{cos}^2 B}{\text{sin}^2 B}\Big)\Big(\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A}\Big)$

Substituting, sin² θ + cos² θ = 1, we get :

⇒ $\Big(\dfrac{1}{{\text{sin}^2 B}}\Big)\Big(\dfrac{1}{\text{sin}^2 A}\Big)$

⇒ cosec² B . cosec² A

Hence, Option 4 is the correct option.

Question 2(i)

Prove that:

(sec A - tan A)² (1 + sin A) = (1 - sin A)

Answer

To prove:

(sec A - tan A)² (1 + sin A) = (1 - sin A)

Solving L.H.S. of the above equation,

$\Rightarrow \text{(sec A - tan A)}^2 \text{(1 + sin A)} \\[1em] \Rightarrow \Big(\dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\Big)^2 \text{(1 + sin A)} \\[1em] \Rightarrow \Big(\dfrac{\text{1 - sin A}}{\text{cos A}}\Big)^2\text{(1 + sin A)} \\[1em] \Rightarrow \dfrac{(1 - \text{sin A})^2\text{(1 + sin A)}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{(1 - \text{sin A})^2\text{(1 + sin A)}}{1 - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{(1 - \text{sin A})^2\text{(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] \Rightarrow \text{1 - sin A}.$

Since, L.H.S. = R.H.S.

Hence, proved that (sec A - tan A)² (1 + sin A) = (1 - sin A).

Question 2(ii)

Prove that :

$\dfrac{\text{cos}^3 A + \text{sin}^3 A}{\text{cos A+ sin A}} + \dfrac{\text{cos}^3 A - \text{sin}^3 A}{\text{cos A - sin A}}$ = 2

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{(\text{cos}^3 A + \text{sin}^3 A)(\text{cos A - sin A}) + (\text{cos}^3 A - \text{sin}^3 A)(\text{cos A + sin A})}{\text{(cos A + sin A)(cos A - sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos}^4 A - \text{cos}^3 A\text{ sin A} + \text{cos A sin}^3 A - \text{ sin}^4 A + \text{cos}^4 A + \text{cos}^3 A\text{ sin A} - \text{ sin}^3 A\text{ cos A} - \text{sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 cos}^4 A - \text{2 sin}^4 A}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2(cos}^4 A - \text{ sin}^4 A)}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2(cos}^2 A - \text{sin}^2 A)\text{(cos}^2 A + \text{sin}^2 A)}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \text{2 (cos}^2 A + \text{sin}^2 A).$

By formula,

cos² A + sin² A = 1

⇒ 2 × 1 = 2.

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cos}^3 A + \text{sin}^3 A}{\text{cos A+ sin A}} + \dfrac{\text{cos}^3 A - \text{sin}^3 A}{\text{cos A - sin A}}$ = 2.

Question 2(iii)

Prove that :

$\dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}}$ = sec A cosec A + 1

Answer

By solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}} \\[1em] \Rightarrow \dfrac{\text{tan A}}{1 - \dfrac{1}{\text{tan A}}} + \dfrac{\dfrac{1}{\text{tan A}}}{\text{1 - tan A}} \\[1em] \Rightarrow \dfrac{\text{tan A}}{\dfrac{\text{tan A - 1}}{\text{tan A}}} + \dfrac{1}{\text{tan A(1 - tan A)}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A}{\text{tan A - 1}} + \dfrac{1}{\text{tan A(1 - tan A)}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A}{\text{tan A - 1}} - \dfrac{1}{\text{tan A(tan A - 1)}} \\[1em] \Rightarrow \dfrac{\text{tan}^3 A - 1}{\text{tan A(tan A - 1)}} \\[1em] \Rightarrow \dfrac{\text{(tan A - 1)(tan}^2 A + \text{ tan A + 1)}}{\text{tan A(tan A - 1)}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A + \text{ tan A + 1}}{\text{tan A}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A}{\text{tan A}} + \dfrac{\text{tan A}}{\text{tan A}} + \dfrac{1}{\text{tan A}} \\[1em] \Rightarrow \text{tan A + 1 + cot A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + 1 + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{sin A cos A + cos}^2 A}{\text{sin A cos A}}.$

By formula,

sin² A + cos² A = 1

$\Rightarrow \dfrac{1 + \text{sin A cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \Rightarrow \dfrac{1}{\text{sin A cos A}} + \dfrac{\text{sin A cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow \text{cosec A sec A} + 1.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{tan A}}{\text{1 - cot A}} + \dfrac{\text{cot A}}{\text{1 - tan A}}$ = sec A cosec A + 1.

Question 2(iv)

Prove that :

$\Big(\text{tan A} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \dfrac{1}{\text{cos A}}\Big)^2 = 2\Big(\dfrac{\text{1 + sin}^2 A}{\text{1 - sin}^2 A}\Big)$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \Big(\text{tan A} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \dfrac{1}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\dfrac{\text{sin A}}{\text{cos A}} - \dfrac{1}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{\text{sin A + 1}}{\text{cos A}}\Big)^2 + \Big(\dfrac{\text{sin A - 1}}{\text{cos A}}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + 1 + \text{2 sin A}}{\text{cos}^2 A} + \dfrac{\text{sin}^2 A + 1 - \text{2 sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + 1 + \text{2 sin A} + \text{sin}^2 A + 1 - \text{2 sin A}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{2(1 + sin}^2 A)}{\text{cos}^2 A}$

By formula,

cos² A = 1 - sin² A

$\Rightarrow 2\Big(\dfrac{1 + \text{sin}^2 A}{1 - \text{sin}^2 A}\Big)$

Since, L.H.S. = R.H.S.

Hence, proved that

$\Big(\text{tan A} + \dfrac{1}{\text{cos A}}\Big)^2 + \Big(\text{tan A} - \dfrac{1}{\text{cos A}}\Big)^2 = 2\Big(\dfrac{\text{1 + sin}^2 A}{\text{1 - sin}^2 A}\Big)$ .

Question 2(v)

Prove that :

2 sin² A + cos⁴ A = 1 + sin⁴ A

Answer

Solving L.H.S. of the equation :

⇒ 2 sin² A + cos⁴ A

⇒ 2 sin² A + (cos² A)²

By formula,

cos² A = 1 - sin² A

⇒ 2 sin² A + (1 - sin² A)²

⇒ 2 sin² A + 1 + sin⁴ A - 2 sin² A

⇒ 1 + sin⁴ A.

Since, L.H.S. = R.H.S.

Hence, proved that 2 sin² A + cos⁴ A = 1 + sin⁴ A.

Question 2(vi)

Prove that :

$\dfrac{\text{sin A - sin B}}{\text{cos A + cos B}} + \dfrac{\text{cos A - cos B}}{\text{sin A + sin B}}$ = 0

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{(sin A - sin B)(sin A + sin B) + (cos A - cos B)(cos A + cos B)}}{\text{(cos A + cos B)(sin A + sin B)}} = 0 \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{sin}^2 B + \text{cos}^2 A - \text{cos}^2 B}{\text{(cos A + cos B)(sin A + sin B)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A -\text{(sin}^2 B + \text{cos}^2 B)}{\text{(cos A + cos B)(sin A + sin B)}}$

By formula,

sin² θ + cos² θ = 1

$\therefore \dfrac{1 - 1}{\text{(cos A + cos B)(sin A + sin B)}} \\[1em] \Rightarrow 0.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin A - sin B}}{\text{cos A + cos B}} + \dfrac{\text{cos A - cos B}}{\text{sin A + sin B}}$ = 0.

Question 2(vii)

Prove that :

(cosec A - sin A)(sec A - cos A) = $\dfrac{1}{\text{tan A + cot A}}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \text{(cosec A - sin A)(sec A - cos A)} \\[1em] \Rightarrow \Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big) \times \Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{1 - sin}^2 A}{\text{sin A}}\Big) \times \Big(\dfrac{\text{1 - cos}^2 A}{\text{cos A}}\Big)$

By formula,

1 - sin² A = cos² A

1 - cos² A = sin² A

$\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \text{cos A sin A}.$

Solving R.H.S. of the equation :

By formula,

sin² θ + cos² θ = 1

$\Rightarrow \dfrac{\text{sin A cos A}}{1} \\[1em] \Rightarrow \text{sin A cos A}.$

Since, L.H.S. = R.H.S.

Hence, proved that (cosec A - sin A)(sec A - cos A) = $\dfrac{1}{\text{tan A + cot A}}$ .

Question 2(viii)

Prove that :

(1 + tan A. tan B)² + (tan A - tan B)² = sec² A sec² B

Answer

By formula,

sec² θ = 1 + tan² θ

Solving L.H.S. of the equation :

⇒ (1 + tan A. tan B)² + (tan A - tan B)²

⇒ 1 + tan² A tan² B + 2 tan A tan B + tan² A + tan² B - 2 tan A tan B

⇒ 1 + tan² A tan² B + tan² A + tan² B

⇒ 1 + tan² A + tan² A tan² B + tan² B

⇒ sec² A + tan² B(tan² A + 1)

⇒ sec² A + tan² B sec² A

⇒ sec² A(1 + tan² B)

⇒ sec² A sec² B.

Since, L.H.S. = R.H.S.

Hence, proved that (1 + tan A. tan B)² + (tan A - tan B)² = sec² A sec² B.

Question 2(ix)

Prove that :

$\dfrac{1}{\text{cos A + sin A - 1}} + \dfrac{1}{\text{cos A + sin A + 1}}$ = cosec A + sec A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{cos A + sin A + 1 + cos A + sin A - 1}}{\text{(cos A + sin A - 1)(cos A + sin A + 1)}} \\[1em] \Rightarrow \dfrac{\text{2(cos A + sin A)}}{\text{(cos A + sin A)}^2 - 1^2} \\[1em] \Rightarrow \dfrac{\text{2(cos A + sin A)}}{\text{cos}^2 A + \text{sin}^2 A + \text{2 cos A sin A} - 1} \\[1em] \Rightarrow \dfrac{\text{2(cos A + sin A)}}{1 - 1 + \text{2 cos A sin A}} \\[1em] \Rightarrow \dfrac{\text{2(cos A + sin A)}}{\text{2 cos A sin A}} \\[1em] \Rightarrow \dfrac{\text{(cos A + sin A)}}{\text{cos A sin A}} \\[1em] \Rightarrow \dfrac{\text{cos A}}{\text{cos A sin A}} + \dfrac{\text{sin A}}{\text{cos A sin A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}} + \dfrac{1}{\text{cos A}}\\[1em] \Rightarrow \text{cosec A + sec A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{1}{\text{cos A + sin A - 1}} + \dfrac{1}{\text{cos A + sin A + 1}}$ = cosec A + sec A.

Question 3

If x cos A + y sin A = m and x sin A - y cos A = n, then prove that :

x² + y² = m² + n²

Answer

To prove:

x² + y² = m² + n²

Substituting value of m and n in R.H.S. of the equation :

= (x cos A + y sin A)² + (x sin A - y cos A)²

= x² cos² A + y² sin² A + 2xy cos A sin A + x² sin² A + y² cos² A - 2xy sin A cos A

= x² cos² A + x² sin² A + y² cos² A + y² sin² A

= x²(sin² A + cos² A) + y²(sin² A + cos² A)

By formula,

sin² A + cos² A = 1

⇒ x² × 1 + y² × 1

⇒ x² + y².

Since, L.H.S. = R.H.S.

Hence, proved that x² + y² = m² + n².

Question 4

If m = a sec A + b tan A and n = a tan A + b sec A, then prove that :

m² - n² = a² - b²

Answer

To prove:

m² - n² = a² - b²

Substituting value of m and n in L.H.S. of the above equation :

= (a sec A + b tan A)² - (a tan A + b sec A)²

= a² sec² A + b² tan² A + 2ab sec A tan A - (a² tan² A + b² sec² A + 2ab sec A tan A)

= a² sec² A - a² tan² A + b²tan² A - b² sec² A + 2ab sec A tan A - 2ab sec A tan A

= a² (sec² A - tan² A) + b² (tan² A - sec² A)

= a² (sec² A - tan² A) - b² (sec² A - tan² A)

By formula,

sec² A - tan² A = 1

⇒ a² × 1 - b² × 1

⇒ a² - b²

Since, L.H.S. = R.H.S.

Hence, proved that m² - n² = a² - b².

Question 5

If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that :

x² + y² + z² = r²

Answer

To prove:

⇒ x² + y² + z² = r²

Substituting value of x, y and z in L.H.S. of the equation :

= (r sin A cos B)² + (r sin A sin B)² + (r cos A)²

= r² sin² A cos² B + r² sin² A sin² B + r² cos² A

= r²sin² A(cos² B + sin² B) + r² cos² A

⇒ r²sin² A + r²cos² A [∵ sin² θ + cos² θ = 1]

⇒ r²(sin² A + cos² A)

⇒ r² × 1 [∵ sin² θ + cos² θ = 1]

⇒ r².

Since, L.H.S. = R.H.S.

Hence, proved that x² + y² + z² = r².

Question 6

If $\dfrac{\text{cos A}}{\text{cos B}} = m \text{ and } \dfrac{\text{cos A}}{\text{sin B}}$ = n,

show that:

(m² + n²) cos² B = n²

Answer

To prove:

(m² + n²) cos² B = n²

Substituting value of m and n in L.H.S. of the above equation :

$\Rightarrow \Big[\Big(\dfrac{\text{cos A}}{\text{cos B}}\Big)^2 + \Big(\dfrac{\text{cos A}}{\text{sin B}}\Big)^2\Big]\text{cos}^2 B \\[1em] \Rightarrow \Big[\dfrac{\text{cos}^2 A}{\text{cos}^2 B} + \dfrac{\text{cos}^2 A}{\text{sin}^2 B}\Big]\text{cos}^2 B \\[1em] \Rightarrow \dfrac{\text{cos}^2 A \text{ sin}^2 B + \text{cos}^2 A \text{cos}^2 B}{\text{ cos}^2 B \text{ sin}^2 B} \times \text{cos}^2 B \\[1em] \Rightarrow \dfrac{\text{cos}^2 A(\text{sin}^2 B + \text{cos}^2 B)}{\text{sin}^2 B}$

By formula,

⇒ sin² B + cos² B = 1

$\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 B} \\[1em] \Rightarrow \Big(\dfrac{\text{cos A}}{\text{sin B}}\Big)^2 \\[1em] \Rightarrow n^2.$

Since, L.H.S. = R.H.S.

Hence, proved that (m² + n²) cos² B = n².

Exercise 21(C)

Question 1(a)

sin² A + sin² (90° - A) is equal to :

-1
1
2
-2

Answer

sin (90° - A) = cos A

To prove:

sin² A + sin² (90° - A)

⇒ sin² A + cos² A

⇒ 1.

Hence, Option 2 is the correct option.

Question 1(b)

In a triangle ABC, sec $\dfrac{C + A}{2}$ is equal to :

cosec $\dfrac{B}{2}$
sec $\dfrac{B}{2}$
cosec $\dfrac{B + A}{2}$
none of these

Answer

In triangle ABC,

By angle sum property of triangle,

⇒ A + B + C = 180°

⇒ A + C = 180° - B

⇒ $\dfrac{A + C}{2} = \dfrac{180° - B}{2}$ ........(1)

Substituting value of $\dfrac{A + C}{2}$ in sec $\dfrac{C + A}{2}$ , we get :

$\Rightarrow \text{sec } \dfrac{C + A}{2} \\[1em] \Rightarrow \text{sec } \dfrac{180° - B}{2} \\[1em] \Rightarrow \text{sec } \Big(90° - \dfrac{B}{2}\Big) \\[1em] \Rightarrow \text{cosec } \dfrac{B}{2}.$

Hence, Option 1 is the correct option.

Question 1(c)

$\dfrac{\text{cot 46°}}{\text{tan 44°}} - 3 \dfrac{\text{sec 20°}}{\text{cosec 70°}}$ + 5 is equal to :

-3
4
3
-4

Answer

Solving,

$\Rightarrow \dfrac{\text{cot 46°}}{\text{tan 44°}} - 3 \dfrac{\text{sec 20°}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow \dfrac{\text{cot (90° - 44°)}}{\text{tan 44°}} - 3 \dfrac{\text{sec (90° - 70°)}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow \dfrac{\text{tan 44°}}{\text{tan 44°}} - 3 \dfrac{\text{cosec 70°}}{\text{cosec 70°}} + 5 \\[1em] \Rightarrow 1 - 3 \times 1 + 5 \\[1em] \Rightarrow 1 - 3 + 5 \\[1em] \Rightarrow 3.$

Hence, Option 3 is the correct option.

Question 1(d)

sin 67° . cos 23° + cos 67° . sin 23° is equal to :

-1
2 sin 67°
2 cos 23°
1

Answer

Solving,

⇒ sin 67° . cos 23° + cos 67° . sin 23°

⇒ sin 67° . cos (90° - 67°) + cos 67° . sin (90° - 67°)

⇒ sin 67° . sin 67° + cos 67° . cos 67°

⇒ sin² 67° + cos² 67°

⇒ 1.

Hence, Option 4 is the correct option.

Question 1(e)

$\dfrac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}}$ is equivalent to :

cos² θ - 1
sin² θ
sin² θ - cos² θ
sin² θ - 1

Answer

Solving,

$\Rightarrow \dfrac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}} \\[1em] \Rightarrow \dfrac{\text{cos θ . sin θ}}{\text{cot θ}} \\[1em] \Rightarrow \dfrac{\text{cos θ . sin θ}}{\dfrac{\text{cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \dfrac{\text{cos θ . sin θ . sin θ}}{\text{cos θ}} \\[1em] \Rightarrow \text{sin}^2 θ.$

Hence, Option 2 is the correct option.

Question 2(i)

Show that :

tan 10° tan 15° tan 75° tan 80° = 1

Answer

Solving L.H.S. of the equation :

⇒ tan 10° tan 15° tan 75° tan 80°

⇒ tan 10° tan 15° tan (90 - 15)° tan (90 - 10)°

By formula,

tan (90° - A) = cot A

⇒ tan 10° tan 15° cot 15° cot 10°

⇒ tan 10° × tan 15° $\times \dfrac{1}{\text{tan 15°}} \times \dfrac{1}{\text{tan 10°}}$

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that tan 10° tan 15° tan 75° tan 80° = 1.

Question 2(ii)

Show that :

sin 42° sec 48° + cos 42° cosec 48° = 2

Answer

Solving L.H.S. of the equation :

⇒ sin 42° sec (90 - 42)° + cos 42° cosec (90 - 42)°

By formula,

sec (90° - A) = cosec A and cosec (90° - A) = sec A.

⇒ sin 42° cosec 42° + cos 42° sec 42°

⇒ sin 42° $\times \dfrac{1}{\text{sin 42°}}$ + cos 42° $\times \dfrac{1}{\text{cos 42°}}$

⇒ 1 + 1

⇒ 2.

Since, L.H.S. = R.H.S.

Hence, proved that sin 42° sec 48° + cos 42° cosec 48° = 2.

Question 3

Express each of the following in terms of angles between 0° and 45° :

(i) sin 59° + tan 63°

(ii) cosec 68° + cot 72°

Answer

(i) Solving,

⇒ sin 59° + tan 63°

⇒ sin (90 - 31)° + tan (90 - 27)°

By formula,

sin (90° - A) = cos A and tan (90° - A) = cot A.

⇒ cos 31° + cot 27°.

Hence, sin 59° + tan 63° = cos 31° + cot 27°.

(ii) Solving,

⇒ cosec 68° + cot 72°

⇒ cosec (90 - 22)° + cot (90 - 18)°

By formula,

cosec (90° - A) = sec A and cot (90° - A) = tan A.

⇒ sec 22° + tan 18°.

Hence, cosec 68° + cot 72° = sec 22° + tan 18°.

Question 4(i)

Show that :

$\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A

Answer

To prove:

$\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A

By formula,

sin (90° - A) = cos A and cos (90° - A) = sin A.

Substituting above values in L.H.S. :

$\Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}$

By formula,

sin² A + cos² A = 1

$\Rightarrow \dfrac{1}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}} \times \dfrac{1}{\text{cos A}} \\[1em] \Rightarrow \text{cosec A sec A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin A}}{\text{sin(90° - A)}} + \dfrac{\text{cos A}}{\text{cos(90° - A)}}$ = sec A cosec A.

Question 4(ii)

Show that :

sin A cos A - $\dfrac{\text{sin A cos (90° - A) cos A}}{\text{sec (90° - A)}} - \dfrac{\text{cos A sin (90° - A) sin A}}{\text{cosec (90° - A)}}$ = 0

Answer

By formula,

cos (90° - A) = sin A, sec (90° - A) = cosec A, cosec (90° - A) = sec A and sin (90° - A) = cos A.

$\Rightarrow \text{sin A cos A} - \dfrac{\text{sin A sin A cos A}}{\text{cosec A}} - \dfrac{\text{cos A cos A sin A}}{\text{sec A}} \\[1em] \Rightarrow \text{sin A cos A} - \dfrac{\text{sin}^2 A \text{cos A}}{\dfrac{1}{\text{sin A}}} - \dfrac{\text{cos}^2 A \text{sin A}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \text{sin A cos A} - \text{sin}^3 A \text{ cos A} - \text{cos}^3 A \text{sin A} \\[1em] \Rightarrow \text{sin A cos A} - \text{sin A cos A}(\text{sin}^2 A + \text{cos}^2 A)$

By formula,

sin² A + cos² A = 1.

$\Rightarrow \text{sin A cos A} - \text{sin A cos A} \\[1em] \Rightarrow 0.$

Since, L.H.S. = R.H.S.

Hence, proved that sin A cos A - $\dfrac{\text{sin A cos (90° - A) cos A}}{\text{sec (90° - A)}} - \dfrac{\text{cos A sin (90° - A) sin A}}{\text{cosec (90° - A)}}$ = 0.

Question 5

For triangle ABC, show that :

(i) sin $\dfrac{A + B}{2} = \text{cos} \dfrac{C}{2}$

(ii) tan $\dfrac{B + C}{2} = \text{cot} \dfrac{A}{2}$

Answer

(i) In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180° [By angle sum property of triangle]

⇒ ∠A + ∠B = 180° - ∠C .........(1)

Given equation,

sin $\dfrac{A + B}{2} = \text{cos} \dfrac{C}{2}$

Substituting value of (A + B) from (1) in L.H.S. of above equation :

$\Rightarrow \text{sin} \dfrac{180° - C}{2} \\[1em] \Rightarrow \text{sin} \Big(90° - \dfrac{C}{2}\Big)$

By formula,

sin(90° - θ) = cos θ

$\therefore \text{cos} \dfrac{C}{2}$ .

Since, L.H.S. = R.H.S.

Hence, proved that sin $\dfrac{A + B}{2} = \text{cos} \dfrac{C}{2}$ .

(ii) In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180° [By angle sum property of triangle]

⇒ ∠B + ∠C = 180° - ∠A .........(1)

Given equation,

tan $\dfrac{B + C}{2} = \text{cot} \dfrac{A}{2}$

Substituting value of (B + C) from (1) in L.H.S. of above equation :

$\Rightarrow \text{tan} \dfrac{180° - A}{2} \\[1em] \Rightarrow \text{tan } \Big(90° - \dfrac{A}{2}\Big)$

By formula,

tan(90° - θ) = cot θ

$\Rightarrow \text{cot} \dfrac{A}{2}$ .

Since, L.H.S. = R.H.S.

Hence, proved that tan $\dfrac{B + C}{2} = \text{cot} \dfrac{A}{2}$ .

Question 6(i)

Evaluate :

3 cos 80° cosec 10° + 2 sin 59° sec 31°

Answer

Solving,

⇒ 3 cos 80° cosec 10° + 2 sin 59° sec 31°

⇒ 3 cos 80° cosec (90 - 80)° + 2 sin 59° sec (90 - 59)°

By formula,

cosec(90° - θ) = sec θ and sec(90° - θ) = cosec θ

⇒ 3 cos 80° sec 80° + 2 sin 59° cosec 59°

⇒ 3 cos 80° $\times \dfrac{1}{\text{cos 80°}}$ + 2 sin 59° $\times \dfrac{1}{\text{sin 59°}}$

⇒ 3 + 2

⇒ 5.

Hence, 3 cos 80° cosec 10° + 2 sin 59° sec 31° = 5.

Question 6(ii)

Evaluate :

$\dfrac{\text{sin 80°}}{\text{cos 10°}}$ + sin 59° sec 31°

Answer

Solving,

$\Rightarrow \dfrac{\text{sin 80°}}{\text{cos 10°}} + \text{sin 59° sec 31°} \\[1em] \Rightarrow \dfrac{\text{sin 80°}}{\text{cos 10°}} + \text{sin 59° sec 31°} \\[1em] \Rightarrow \dfrac{\text{sin (90 - 10)°}}{\text{cos 10°}} + \text{sin 59° sec (90 - 59)°}$

By formula,

sin(90° - θ) = cos θ and sec(90° - θ) = cosec θ

$\Rightarrow \dfrac{\text{cos 10°}}{\text{cos 10°}} + \text{sin 59° cosec 59°} \\[1em] \Rightarrow 1 + \text{sin 59°} \times \dfrac{1}{\text{sin 59°}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.$

Hence, $\dfrac{\text{sin 80°}}{\text{cos 10°}}$ + sin 59° sec 31° = 2.

Question 6(iii)

Evaluate :

tan(55° - A) - cot(35° + A)

Answer

Solving,

⇒ tan(55° - A) - cot(35° + A)

⇒ tan[90° - (35° + A)] - cot(35° + A)

By formula,

tan(90° - θ) = cot θ

⇒ cot(35° + A) - cot(35° + A)

⇒ 0.

Hence, tan(55° - A) - cot(35° + A) = 0

Question 6(iv)

Evaluate :

cosec (65° + A) - sec (25° - A)

Answer

Solving,

⇒ cosec (65° + A) - sec (25° - A)

⇒ cosec [90° - (25° - A)] - sec (25° - A)

By formula,

cosec(90° - θ) = sec θ

⇒ sec (25° - A) - sec (25° - A)

⇒ 0.

Hence, cosec (65° + A) - sec (25° - A) = 0.

Question 6(v)

Evaluate :

$2\dfrac{\text{tan 57°}}{\text{cot 33°}} - \dfrac{\text{cot 70°}}{\text{tan 20°}} - \sqrt{2}$ cos 45°

Answer

Solving,

$\Rightarrow 2\dfrac{\text{tan 57°}}{\text{cot 33°}} - \dfrac{\text{cot 70°}}{\text{tan 20°}} - \sqrt{2}\text{cos 45°} \\[1em] \Rightarrow 2\dfrac{\text{tan 57°}}{\text{cot (90° - 57°)}} - \dfrac{\text{cot 70°}}{\text{tan (90° - 70°)}} - \sqrt{2}\times \dfrac{1}{\sqrt{2}}$

By formula,

tan(90° - θ) = cot θ and cot(90° - θ) = tan θ

$\Rightarrow 2 \times \dfrac{\text{tan 57°}}{\text{tan 57°}} - \dfrac{\text{cot 70°}}{\text{cot 70°}} - 1 \\[1em] \Rightarrow 2 - 1 - 1 \\[1em] \Rightarrow 0.$

Hence, $2\dfrac{\text{tan 57°}}{\text{cot 33°}} - \dfrac{\text{cot 70°}}{\text{tan 20°}} - \sqrt{2}$ cos 45° = 0.

Question 6(vi)

Evaluate :

$\dfrac{\text{cot}^2 41°}{\text{tan}^2 49°} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 15°}$

Answer

Solving,

$\Rightarrow \dfrac{\text{cot}^2 41°}{\text{tan}^2 49°} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 15°} \\[1em] \Rightarrow \dfrac{\text{cot}^2 41°}{\text{tan}^2 (90° - 41°)} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 (90° - 75°)}$

By formula,

tan (90° - θ) = cot θ and cos (90° - θ) = sin θ

$\Rightarrow \dfrac{\text{cot}^2 41°}{\text{cot}^2 41°} - 2\dfrac{\text{sin}^2 75°}{\text{sin}^2 75°} \\[1em] \Rightarrow 1 - 2 \\[1em] \Rightarrow -1.$

Hence, $\dfrac{\text{cot}^2 41°}{\text{tan}^2 49°} - 2\dfrac{\text{sin}^2 75°}{\text{cos}^2 15°}$ = -1

Question 6(vii)

$\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 30°$

Answer

Solving,

$\Rightarrow \dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 30° \\[1em] \Rightarrow \dfrac{\text{cos 70°}}{\text{sin (90° - 70°)}} + \dfrac{\text{cos 59°}}{\text{sin (90° - 59°)}} - \text{8 sin}^2 30°$

By formula,

sin (90° - θ) = cos θ.

$\Rightarrow \dfrac{\text{cos 70°}}{\text{cos 70°}} + \dfrac{\text{cos 59°}}{\text{cos 59°}} - 8 \times \Big(\dfrac{1}{2}\Big)^2 \\[1em] \Rightarrow 1 + 1 - 8\times \dfrac{1}{4} \\[1em] \Rightarrow 2 - 2 \\[1em] \Rightarrow 0.$

Hence, $\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 30°$ = 0.

Question 7

A triangle ABC is right angled at B; find the value of $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$

Answer

In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ ∠A + ∠C = 90°

⇒ ∠A = 90° - ∠C.

Substituting value of A in $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ we get,

$\Rightarrow \dfrac{\text{sec (90° - C). cosec C - tan (90° - C). cot C}}{\text{sin 90°}}$

By formula,

tan (90° - C) = cot c, sec (90° - C) = cosec C and cosec² C - cot² C = 1.

$\Rightarrow \dfrac{\text{cosec C. cosec C - cot C. cot C}}{1} \\[1em] \Rightarrow \text{cosec}^2 C - \text{cot}^2 C \\[1em] \Rightarrow 1.$

Hence, $\dfrac{\text{sec A. cosec C - tan A. cot C}}{\text{sin B}}$ = 1.

Question 8(i)

Find value of x, if :

sin x = sin 60° cos 30° + cos 60° sin 30°

Answer

Solving,

$\Rightarrow \text{sin x} = \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] \Rightarrow \text{sin x} = \dfrac{3}{4} + \dfrac{1}{4} \\[1em] \Rightarrow \text{sin x} = \dfrac{4}{4} \\[1em] \Rightarrow \text{sin x} = 1 \\[1em] \Rightarrow \text{sin x} = \text{sin 90°} \\[1em] \Rightarrow x = 90°.$

Hence, x = 90°.

Question 8(ii)

Find value of x, if :

cos x = cos 60° cos 30° - sin 60° sin 30°

Answer

Solving,

$\Rightarrow \text{cos x} = \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2} \times \dfrac{1}{2} \\[1em] \Rightarrow \text{cos x} = \dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4} \\[1em] \Rightarrow \text{cos x} = 0 \\[1em] \Rightarrow \text{cos x} = \text{cos 90°} \\[1em] \Rightarrow x = 90°.$

Hence, x = 90°.

Question 8(iii)

Find value of x, if :

tan x = $\dfrac{\text{tan 60° - tan 30°}}{\text{1 + tan 60° tan 30°}}$

Answer

Solving,

$\Rightarrow \text{tan x} = \dfrac{\text{tan 60° - tan 30°}}{\text{1 + tan 60° tan 30°}} \\[1em] \Rightarrow \text{tan x} = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \dfrac{1}{\sqrt{3}}} \\[1em] \Rightarrow \text{tan x} = \dfrac{\dfrac{3 - 1}{\sqrt{3}}}{1 + 1} \\[1em] \Rightarrow \text{tan x} = \dfrac{\dfrac{2}{\sqrt{3}}}{2} \\[1em] \Rightarrow \text{tan x} = \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow \text{tan x} = \text{tan 30°} \\[1em] \Rightarrow x = 30°.$

Hence, x = 30°.

Question 8(iv)

Find value of x, if :

sin 3x = 2 sin 30° cos 30°

Answer

Solving,

⇒ sin 3x = 2 sin 30° cos 30°

⇒ sin 3x = 2 $\times \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}$

⇒ sin 3x = $\dfrac{\sqrt{3}}{2}$

⇒ sin 3x = sin 60°

⇒ 3x = 60°

⇒ x = $\dfrac{60°}{3}$ = 20°.

Hence, x = 20°.

Question 8(v)

Find value of x, if :

cos(2x - 6°) = cos² 30° - cos² 60°

Answer

Solving,

⇒ cos(2x - 6°) = cos² 30° - cos² 60°

⇒ cos(2x - 6°) = $\Big(\dfrac{\sqrt{3}}{2}\Big)^2 - \Big(\dfrac{1}{2}\Big)^2$

⇒ cos(2x - 6°) = $\dfrac{3}{4} - \dfrac{1}{4}$

⇒ cos(2x - 6°) = $\dfrac{2}{4}$

⇒ cos(2x - 6°) = $\dfrac{1}{2}$

⇒ cos(2x - 6°) = cos 60°

⇒ 2x - 6° = 60°

⇒ 2x = 66°

⇒ x = 33°.

Hence, x = 33°.

Question 9

In each case given below, find the value of angle A, where 0° ≤ A ≤ 90°.

(i) sin (90° - 3A). cosec 42° = 1

(ii) cos (90° - A). sec 77° = 1

Answer

(i) Given,

⇒ sin (90° - 3A). cosec 42° = 1

⇒ sin (90° - 3A) × $\dfrac{1}{\text{sin 42°}}$ = 1

⇒ sin (90° - 3A) = sin 42°

⇒ 90° - 3A = 42°

⇒ 3A = 90° - 42°

⇒ 3A = 48°

⇒ A = 16°.

Hence, A = 16°.

(ii) Given,

⇒ cos (90° - A). sec 77° = 1

⇒ cos (90° - A) × $\dfrac{1}{\text{cos 77°}}$ = 1

⇒ cos (90° - A) = cos 77°

⇒ 90° - A = 77°

⇒ A = 90° - 77°

⇒ A = 13°.

Hence, A = 13°.

Question 10(i)

Prove that :

$\dfrac{\text{cos(90° - θ) cos θ}}{\text{cot θ}}$ = 1 - cos² θ

Answer

By formula,

cos (90° - θ) = sin θ.

Solving L.H.S. of the equation

$\Rightarrow \dfrac{\text{cos(90° - θ) cos θ}}{\text{cot θ}} \\[1em] \Rightarrow \dfrac{\text{sin θ cos θ}}{\dfrac{\text{cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \text{sin}^2 θ \\[1em] \Rightarrow \text{1 - cos}^2 θ.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cos(90° - θ) cos θ}}{\text{cot θ}}$ = 1 - cos² θ.

Question 11

Evaluate :

$\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°}$

Answer

Solving,

$\Rightarrow \dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°} \\[1em] \Rightarrow \dfrac{\text{sin 35° cos (90° - 35°) + cos 35° sin (90° - 35°)}}{\text{cosec}^2 10° - \text{tan}^2 (90° - 10°)}$

By formula,

sin(90° - θ) = cos θ, cos(90° - θ) = sin θ and tan(90° - θ) = cot θ.

$\Rightarrow \dfrac{\text{sin 35° sin 35° + cos 35° cos 35°}}{\text{cosec}^2 10° - \text{cot}^2 10°} \\[1em] \Rightarrow \dfrac{\text{sin}^2 35° + \text{cos}^2 35°}{\text{cosec}^2 10° - \text{cot}^2 10°}$

By formula,

sin² θ + cos² θ = 1 and cosec² θ - cot² θ = 1.

$\Rightarrow \dfrac{1}{1}$

⇒ 1.

Hence, $\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°}$ = 1.

Question 12

Evaluate :

sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30°

Answer

Solving,

⇒ sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30°

⇒ sin² 34° + sin² (90° - 34°) + 2 tan 18° tan (90° - 18°) - cot² 30°

By formula,

sin(90° - A) = cos A and tan(90° - A) = cot A

⇒ sin² 34° + cos² 34° + 2 tan 18° cot 18° - cot² 30°

By formula,

sin² θ + cos² θ = 1

⇒ 1 + 2 tan 18° $\times \dfrac{1}{\text{tan 18°}} - (\sqrt{3})^2$

⇒ 1 + 2 - 3

⇒ 3 - 3

⇒ 0.

Hence, sin² 34° + sin² 56° + 2 tan 18° tan 72° - cot² 30° = 0.

Question 13

Evaluate :

cosec² 57° - tan² 33° + cos 44° cosec 46° - $\sqrt{2}$ cos 45° - tan² 60°

Answer

Solving,

⇒ cosec² 57° - tan² 33° + cos 44° cosec 46° - $\sqrt{2}$ cos 45° - tan² 60°

⇒ cosec² 57° - tan² (90° - 57°) + cos 44° cosec (90° - 44°) - $\sqrt{2} \times \dfrac{1}{\sqrt{2}} - (\sqrt{3})^2$

By formula,

tan(90° - A) = cot A and cosec(90° - A) = sec A

⇒ cosec² 57° - cot² 57° + cos 44° sec 44° - 1 - 3

By formula,

cosec² A - cot² A = 1

⇒ 1 + cos 44° $\times \dfrac{1}{\text{cos 44°}}$ - 1 - 3

⇒ 1 + 1 - 1 - 3

⇒ -2.

Hence, cosec² 57° - tan² 33° + cos 44° cosec 46° - $\sqrt{2}$ cos 45° - tan² 60° = -2.

Exercise 21(D)

Question 1(a)

If 2 sin θ = $\sqrt{3}$ , then :

2 cos θ + 1 = 0
1 - sin θ = 0
cos θ = $\sqrt{3}$
2 cos θ = 1

Answer

Given,

⇒ 2 sin θ = $\sqrt{3}$

⇒ sin θ = $\dfrac{\sqrt{3}}{2}$

⇒ sin θ = sin 60°

⇒ θ = 60°

⇒ cos θ = cos 60°

⇒ 2 cos θ = 2 cos 60°

⇒ 2 cos θ = $2 \times \dfrac{1}{2}$

⇒ 2 cos θ = 1.

Hence, Option 4 is the correct option.

Question 1(b)

If cos 62° 40' = x, the value of 1 + sin² 27° 20' is :

1 + x²
1 - x²
x² - 1
1 + x

Answer

Given,

⇒ cos 62° 40' = x

⇒ cos (90° - 27° 20') = x

⇒ sin 27° 20' = x

Substituting value of sin 27° 20' in 1 + sin² 27° 20', we get :

⇒ 1 + sin² 27° 20' = 1 + x².

Hence, Option 1 is the correct option.

Question 1(c)

sin θ = 0.5 implies tan θ + cot θ is equal to :

2 : $\sqrt{3}$
$\sqrt{3}$ : 4
4 : $\sqrt{3}$
$\sqrt{3}$ : 2

Answer

Given,

⇒ sin θ = 0.5

⇒ sin θ = $\dfrac{5}{10}$

⇒ sin θ = $\dfrac{1}{2}$

⇒ sin θ = sin 30°

⇒ θ = 30°.

Substituting value of θ in tan θ + cot θ, we get :

⇒ tan θ + cot θ

⇒ tan 30° + cot 30°

⇒ $\dfrac{1}{\sqrt{3}} + \sqrt{3}$

⇒ $\dfrac{1 + 3}{\sqrt{3}}$

⇒ $\dfrac{4}{\sqrt{3}}$

⇒ 4 : $\sqrt{3}$ .

Hence, Option 3 is the correct option.

Question 1(d)

If cos 63° 36' = 0.4446, the value of sin 26° 24' is :

1
1 + 0.4446
0.4446
1 - 0.4446

Answer

Given,

⇒ cos 63° 36' = 0.4446

⇒ cos (90° - 26° 24') = 0.4446

⇒ sin 26° 24' = 0.4446

Hence, Option 3 is the correct option.

Question 2

Use tables to find sine of :

(i) 21°

(ii) 34° 42'

(iii) 47° 32'

(iv) 62° 57'

Answer

(i) From table,

sin 21° = 0.3584

Hence, sin 21° = 0.3584

(ii) From table,

sin 34° 42' = 0.5693

Hence, sin 34° 42' = 0.5693

(iii) From table,

sin 47° 30' = 0.7373

Difference of 2' = 0.0004 [To add]

∴ sin 47° 32' = 0.7373 + 0.0004 = 0.7377

Hence, sin 47° 32' = 0.7377

(iv) From table,

sin 62° 54' = 0.8902

Difference of 3' = 0.0004 [To add]

∴ sin 62° 57' = 0.8902 + 0.0004 = 0.8906

Hence, sin 62° 57' = 0.8906

Question 3

Use tables to find cosine of :

(i) 2° 4'

(ii) 8° 12'

(iii) 26° 32'

(iv) 65° 41'

Answer

(i) From table,

cos 2° 6' = 0.9993

Difference of 2' = 0 [To add]

∴ cos 2° 4' = 0.9993 + 0 = 0.9993

Hence, 2° 4' = 0.9993

(ii) From table,

cos 8° 12' = 0.9898

Hence, 8° 12' = 0.9898

(iii) From table,

cos 26° 30' = 0.8949

Difference of 2' = 0.0003 [To subtract]

∴ cos 26° 32' = 0.8949 - 0.0003 = 0.8946

Hence, 26° 32' = 0.8946

(iv) From table,

cos 65° 42' = 0.4115

Difference of 1' = 0.0003 [To add]

∴ cos 65° 41' = 0.4115 + 0.0003 = 0.4118

Hence, 65° 41' = 0.4118

Question 4

Use trigonometric tables to find tangent of :

(i) 37°

(ii) 42° 18'

Answer

(i) From table,

tan 37° = 0.7536

Hence, tan 37° = 0.7536

(ii) From table,

tan 42° 18' = 0.9099

Hence, tan 42° 18' = 0.9099

Question 5

Use tables to find the acute angle θ, if the value of sin θ is :

(i) 0.4848

(ii) 0.3827

Answer

(i) Let sin θ = 0.4848

From table,

sin 29° = 0.4848

⇒ θ = 29°.

Hence, θ = 29°.

(ii) Let sin θ = 0.3827

From graph,

sin 22° 30' = 0.3827

θ = 22° 30'.

Hence, θ = 22° 30'.

Question 6

Use tables to find the acute angle θ, if the value of cos θ is :

(i) 0.9848

(ii) 0.9574

Answer

(i) Let cos θ = 0.9848

From graph,

cos 10° = 0.9848

θ = 10°.

Hence, θ = 10°.

(ii) Let cos θ = 0.9574

From graph,

cos 16° 48' = 0.9573

cos θ - cos 16° 48' = 0.9574 - 0.9573 = 0.0004

From table,

Difference of 1' = 0.0004 [To subtract]

∴ θ = 16° 48' - 1' = 16° 47'.

Hence, θ = 16° 47'.

Question 7

Use tables to find the acute angle θ, if the value of tan θ is :

(i) 0.2419

(ii) 0.4741

Answer

(i) Let tan θ = 0.2419

From table,

tan 13° 36' = 0.2419

θ = 13° 36'.

Hence, θ = 13° 36'.

(ii) Let tan θ = 0.4741

From table,

tan 25° 18' = 0.4727

tan θ - tan 25° 18' = 0.4741 - 0.4727 = 0.0014

From table,

Difference of 4' = 0.0014

∴ θ = 25° 18' + 4' = 25° 22'.

Hence, θ = 25° 22'.

Test Yourself

Question 1(a)

For the given figure, which of the following conditions is true :

cot γ < cot β
cot γ > cot α
cot β > cot α
cot β < cot γ

Answer

From figure,

cot α = $\dfrac{OA}{AB}$

cot β = $\dfrac{OA}{AC}$

cot γ = $\dfrac{OA}{AD}$

Since, AB < AC < AD

∴ cot γ < cot β < cot α

Hence, Option 1 is the correct option.

Question 1(b)

If sin A = $\dfrac{4}{5}$ , the value of $\sqrt{\dfrac{1 - \text{cos (90° - A)}}{1 + \text{cos (90° - A)}}}$ is :

1
3
$\dfrac{1}{2}$
$\dfrac{1}{3}$

Answer

Given,

sin A = $\dfrac{4}{5}$

We need to find the value of:

$\sqrt{\dfrac{1 - \text{cos (90° - A)}}{1 + \text{cos (90° - A)}}}$

Solving,

$\Rightarrow \sqrt{\dfrac{1 - \text{cos (90° - A)}}{1 + \text{cos (90° - A)}}} \\[1em] \Rightarrow \sqrt{\dfrac{1 - \text{sin A}}{1 + \text{sin A}}} \\[1em] \Rightarrow \sqrt{\dfrac{1 - \dfrac{4}{5}}{1 + \dfrac{4}{5}}} \\[1em] \Rightarrow \sqrt{\dfrac{\dfrac{5 - 4}{5}}{\dfrac{5 + 4}{5}}} \\[1em] \Rightarrow \sqrt{\dfrac{\dfrac{1}{5}}{\dfrac{9}{5}}} \\[1em] \Rightarrow \sqrt{\dfrac{1 \times 5}{5 \times 9}} \\[1em] \Rightarrow \sqrt{\dfrac{1}{9}} \\[1em] \Rightarrow \dfrac{1}{3}.$

Hence, Option 4 is the correct option.

Question 1(c)

(1 + tan² A) × cos² A - 1 is :

1
-2
0
2

Answer

Solving,

$\Rightarrow (1 + \text{tan}^2 A) \times \text{cos}^2 A - 1 \\[1em] \Rightarrow \Big(1 + \dfrac{\text{sin}^2 A}{\text{cos}^2 A}\Big) \times \text{cos}^2 A - 1 \\[1em] \Rightarrow \Big(\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}\Big) \times \text{cos}^2 A - 1 \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 A} \times \text{cos}^2 A - 1 \\[1em] \Rightarrow 1 - 1 \\[1em] \Rightarrow 0.$

Hence, Option 3 is the correct option.

Question 1(d)

$\dfrac{\text{sin A tan A}}{1 - \text{cos}^2 A}$ is equal to :

cos A
sec A
cosec A
sin A

Answer

Solving,

$\Rightarrow \dfrac{\text{sin A tan A}}{1 - \text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin A tan A}}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{tan A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin A}}{\text{cos A}}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{sin A} \times \text{cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} \\[1em] \Rightarrow \text{sec A}.$

Hence, Option 2 is the correct option.

Question 1(e)

For acute angles α and β, if α > β, then :

A : cos α > cos β

B : tan α > tan β

C : sin α > sin β

D : cot α > cot β

Which of the above statement/statements is/are true :

A and B
B and C
C and D
D and A

Answer

For acute angles,

Greater the angle, greater is the sine and tangent of the angle.

Since,

α > β

∴ tan α > tan β

∴ sin α > sin β.

Hence, Option 2 is the correct option.

Question 1(f)

If M = x cos A + y sin A and N = x sin A - y cos A, then M² + N² is :

x² - y²
y² - x²
x² + y²
(x + y)²

Answer

Substituting value of M and N in M² + N², we get :

⇒ (x cos A + y sin A)² + (x sin A - y cos A)²

⇒ x² cos² A + y² sin² A + 2xy sin A cos A + x² sin² A + y² cos² A - 2xy sin A cos A

⇒ x² cos² A + y² sin² A + x² sin² A + y² cos² A

⇒ x² cos² A + x² sin² A + y² sin² A + y² cos² A

⇒ x² (sin² A + cos² A) + y² (sin² A + cos² A)

Substituting, sin² A + cos² A = 1, we get :

⇒ x² × 1 + y² × 1

⇒ x² + y².

Hence, Option 3 is the correct option.

Question 1(g)

$\dfrac{\text{tan 60° - tan 30°}}{1 + \text{tan 60° tan 30°}}$ is equal to :

tan 30°
tan 45°
tan 60°
none

Answer

Solving,

$\Rightarrow \dfrac{\text{tan 60° - tan 30°}}{1 + \text{tan 60° tan 30°}} \\[1em] \Rightarrow \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3} \times \dfrac{1}{\sqrt{3}}} \\[1em] \Rightarrow \dfrac{\dfrac{3 - 1}{\sqrt{3}}}{1 + 1} \\[1em] \Rightarrow \dfrac{\dfrac{2}{\sqrt{3}}}{2} \\[1em] \Rightarrow \dfrac{2}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow \text{tan 30°}.$

Hence, Option 1 is the correct option.

Question 1(h)

$3\dfrac{\text{sin 47°}}{\text{cos 43°}} - \dfrac{\text{tan 50°}}{\text{cot 40°}} - \text{2 sin}^2 45°$ is equal to :

0
-1
1
2

Answer

Solving,

$\Rightarrow 3\dfrac{\text{sin 47°}}{\text{cos 43°}} - \dfrac{\text{tan 50°}}{\text{cot 40°}} - \text{2 sin}^2 45° \\[1em] \Rightarrow 3\dfrac{\text{sin 47°}}{\text{cos (90° - 47°)}} - \dfrac{\text{tan 50°}}{\text{cot (90° - 50°)}} - \text{2 sin}^2 45° \\[1em] \Rightarrow 3\dfrac{\text{sin 47°}}{\text{sin 47°}} - \dfrac{\text{tan 50°}}{\text{tan 50°}} - 2 \times \Big(\dfrac{1}{\sqrt{2}}\Big)^2 \\[1em] \Rightarrow 3 \times 1 - 1 - 2 \times \dfrac{1}{2} \\[1em] \Rightarrow 3 - 1 - 1 \\[1em] \Rightarrow 1.$

Hence, Option 3 is the correct option.

Question 1(i)

If sin 2x = 2 sin 45° cos 45°; the value of x is :

45°
90°
30°
60°

Answer

Solving,

⇒ sin 2x = 2 sin 45° cos 45°

⇒ sin 2x = $2 \times \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}}$

⇒ sin 2x = $2 \times \dfrac{1}{2}$

⇒ sin 2x = 1

⇒ sin 2x = sin 90°

⇒ 2x = 90°

⇒ x = $\dfrac{90°}{2}$ = 45°.

Hence, Option 1 is the correct option.

Question 1(j)

For acute angle θ, sec²θ = 1 - tan²θ

Assertion(A): sec²θ = 1 - tan²θ is not a trigonometric identity.

Reason(R): For an acute angle θ, the trigonometric equation is an identity, if it is satisfied for every value of angle θ.

A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.

Answer

The valid trigonometric identity;

sec²θ = 1 + tan²θ

So, assertion (A) is true.

A trigonometric identity is an equation that holds true for all values of the variable (in this case, the angle θ) within its domain.

So, reason (R) is true.

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Question 1(k)

(1 + tan² A)(1 - sin A)(1 + sin A)

Assertion(A): The value of given trigonometric expression is 0.

Reason(R): The given expression is equal to sec² A.cos² A

A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.

Answer

Given,

⇒ (1 + tan² A)(1 - sin A)(1 + sin A)

⇒ (1 + tan² A)(1 - sin² A)

⇒ (1 + tan² A)cos² A

⇒ sec² A.cos² A

So, reason (R) are true.

⇒ $\dfrac{1}{\text{cos}^2 A}$ .cos² A

⇒ 1.

So, assertion (A) is false.

∴ A is false, R is true.

Hence, option 2 is the correct option.

Question 1(l)

cos A = $\dfrac{\sqrt{3}}{2}$ and sin B = $\dfrac{1}{2}$

Assertion (A): tan (A + B) = $\sqrt{3}$

Reason (R): cos A = $\dfrac{\sqrt{3}}{2}$ ⇒ A = 30°

sin B = $\dfrac{1}{2}$ ⇒ B = 30°

A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.

Answer

Given, cos A = $\dfrac{\sqrt{3}}{2}$ and sin B = $\dfrac{1}{2}$

⇒ cos A = cos 30° and sin B = sin 30°

⇒ A = 30° and B = 30°

So, reason (R) is true.

tan(A + B) = tan(30° + 30°) = tan 60° = $\sqrt{3}$

Thus, tan (A + B) = $\sqrt{3}$

So, assertion (A) is true.

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Question 1(m)

x = (cosec A + cot A)(1 - cos A)

Assertion (A): x = sin A

Reason (R): x = $\Big(\dfrac{1}{\text{sin A}} + \dfrac{\text{cos A}}{\text{sin A}}\Big)(1 - \text{cos A}) = \dfrac{\text{sin}^2 A}{\text{sin A}} = \text{sin A}$

A is true, R is false.
A is false, R is true.
Both A and R are true and R is correct reason for A.
Both A and R are true and R is incorrect reason for A.

Answer

Given, x = (cosec A + cot A)(1 - cos A)

$\Rightarrow x = \Big(\dfrac{1}{\text{sin A}} + \dfrac{\text{cos A}}{\text{sin A}}\Big)(1 - \text{cos A})\\[1em] = \Big(\dfrac{1 + \text{cos A}}{\text{sin A}}\Big)(1 - \text{cos A})\\[1em] = \dfrac{(1 + \text{cos A})(1 - \text{cos A})}{\text{sin A}}\\[1em] = \Big(\dfrac{1 - \text{cos}^2 A}{\text{sin A}}\Big)\\[1em] = \Big(\dfrac{\text{sin}^2 A}{\text{sin A}}\Big)\\[1em] = \text{sin A}.$

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Question 1(n)

$\dfrac{1 - \text{sin A}}{\text{cos A}}$ = sec A - tan A

Statement (1): $\dfrac{1 - \text{sin A}}{\text{cos A}}$ = sec A - tan A

$\Rightarrow \dfrac{1 + \text{sin A}}{\text{cos A}}$ = sec A + tan A

Statement (2): $\dfrac{1 - \text{sin A}}{\text{cos A}} - \dfrac{1 + \text{sin A}}{\text{cos A}}$ = 2sec A

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given, $\dfrac{1 - \text{sin A}}{\text{cos A}}$ = sec A - tan A

Taking L.H.S.

$\Rightarrow \dfrac{1 - \text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \text{sec A} - \text{tan A}$

As, L.H.S. = R.H.S., so equation is true.

Now, $\dfrac{1 + \text{sin A}}{\text{cos A}}$ = sec A + tan A

Taking L.H.S.

$\Rightarrow \dfrac{1 + \text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\\[1em] \Rightarrow \text{sec A} + \text{tan A}$

As, L.H.S. = R.H.S., so equation is true.

So, statement 1 is true.

Now, $\dfrac{1 - \text{sin A}}{\text{cos A}} - \dfrac{1 + \text{sin A}}{\text{cos A}}$ = 2sec A

Solving L.H.S.

$\Rightarrow \dfrac{1 - \text{sin A}}{\text{cos A}} - \dfrac{1 + \text{sin A}}{\text{cos A}}$

$\Rightarrow \dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}} - \Big(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big)$

⇒ (sec A - tan A) - (sec A + tan A)

⇒ sec A - tan A - sec A - tan A

⇒ -2tan A.

As, L.H.S. ≠ R.H.S.

So, statement 2 is false.

∴ Statement 1 is true and statement 2 is false.

Hence, option 3 is the correct option.

Question 1(o)

$\text{cos}^2 θ + \dfrac{1}{1 + {\text{cot}^2 θ}}$ = x

Statement (1): x = 1

Statement (2): x = $\text{cos}^2 θ + \dfrac{1}{{\text{cosec}^2 θ}}$ = cos² θ + sin² θ

Both the statement are true.
Both the statement are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given, $\text{cos}^2 θ + \dfrac{1}{1 + {\text{cot}^2 θ}}$ = x

$\Rightarrow x = \text{cos}^2 θ + \dfrac{1}{1 + {\text{cot}^2 θ}}\\[1em] \Rightarrow x = \text{cos}^2 θ + \dfrac{1}{{\text{cosec}^2 θ}}\\[1em] \Rightarrow x = \text{cos}^2 θ + \text{sin}^2 θ\\[1em] \Rightarrow x = 1$

∴ Both the statements are true.

Hence, option 1 is the correct option.

Question 1(p)

$\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \dfrac{1}{ \text{cos}^2 θ}$ = x

Statement (1): x = 1

Statement (2): x = $\dfrac{\text{sin}^2 θ - 1}{{\text{cos}^2 θ}} = \dfrac{-\text{cos}^2 θ}{{\text{cos}^2 θ}}$ = -1

Both the statement are true.
Both the statement are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given, $\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \dfrac{1}{ \text{cos}^2 θ}$ = x

$\Rightarrow x = \dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} - \dfrac{1}{ \text{cos}^2 θ}\\[1em] \Rightarrow x = \dfrac{\text{sin}^2 θ - 1}{\text{cos}^2 θ}\\[1em] \Rightarrow x = \dfrac{\text{sin}^2 θ - (\text{sin}^2 θ + \text{cos}^2 θ)}{\text{cos}^2 θ} \\[1em] \Rightarrow x = \dfrac{\text{sin}^2 θ - \text{sin}^2 θ - \text{cos}^2 θ}{\text{cos}^2 θ}\\[1em] \Rightarrow x = \dfrac{- \text{cos}^2 θ}{\text{cos}^2 θ}\\[1em] \Rightarrow x = -1$

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Question 1(q)

sin A = cos A and tan² A + cot² A + 2

Statement (1): A = 45°

tan² A + cot² A + 2 = 4

Statement (2): sin A = cos A ⇒ A = 45°

Both the statement are true.
Both the statement are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given, sin A = cos A

⇒ $\dfrac{\text{sin A}}{\text{cos A}}$ = 1

⇒ tan A = 1

⇒ tan A = tan 45°

⇒ A = 45°

So, statement 2 is true.

Substituting value of A in tan² A + cot² A + 2, we get :

⇒ tan² A + cot² A + 2 = tan² 45° + cot² 45° + 2

= (1)² + (1)² + 2

= 1 + 1 + 2

= 4.

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Question 2(i)

Prove the following identities :

$\dfrac{1}{\text{cos A + sin A}} + \dfrac{1}{\text{cos A - sin A}} = \dfrac{\text{2 cos A}}{\text{2 cos}^2 A - 1}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1}{\text{cos A + sin A}} + \dfrac{1}{\text{cos A - sin A}} \\[1em] \Rightarrow \dfrac{\text{cos A - sin A + cos A + sin A}}{\text{cos}^2 A - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 cos A}}{\text{cos}^2 A - \text{sin}^2 A}$

By formula,

sin² A = 1 - cos² A

$\Rightarrow \dfrac{\text{2 cos A}}{\text{cos}^2 A - (1 - \text{cos}^2 A)} \\[1em] \Rightarrow \dfrac{\text{2 cos A}}{\text{2 cos}^2 A - 1}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{1}{\text{cos A + sin A}} + \dfrac{1}{\text{cos A - sin A}} = \dfrac{\text{2 cos A}}{\text{2 cos}^2 A - 1}$ .

Question 2(ii)

Prove the following identities :

$1 - \dfrac{\text{sin}^2 A}{\text{1 + cos A}} = \text{cos A}$

Answer

By formula,

sin² A = 1 - cos² A

Solving L.H.S. of the equation :

$\Rightarrow 1 - \dfrac{1 - \text{cos}^2 A}{\text{1 + cos A}} \\[1em] \Rightarrow 1 - \dfrac{\text{(1 - cos A)(1 + cos A)}}{\text{1 + cos A}} \\[1em] \Rightarrow 1 - (\text{1 - cos A}) \\[1em] \Rightarrow 1 - 1 + \text{cos A} \\[1em] \Rightarrow \text{cos A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $1 - \dfrac{\text{sin}^2 A}{\text{1 + cos A}} = \text{cos A}$ .

Question 2(iii)

Prove the following identities :

$\dfrac{\text{cos A}}{\text{1 + sin A}} + \text{tan A = sec A}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{cos A}}{\text{1 + sin A}} + \text{tan A}\\[1em] \Rightarrow \dfrac{\text{cos A}}{\text{1 + sin A}} + \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A + \text{sin A(1 + sin A)}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A + \text{sin A + sin}^2 A}{\text{cos A(1 + sin A)}}$

By formula,

cos² A + sin² A = 1

$\Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} \\[1em] \Rightarrow \text{sec A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cos A}}{\text{1 + sin A}} + \text{tan A = sec A}$ .

Question 2(iv)

Prove the following identities :

$\dfrac{\text{sin A}}{\text{1 - cos A}} - \text{cot A = cosec A}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{sin A}}{\text{1 - cos A}} - \text{cot A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{1 - cos A}} - \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{cos A(1 - cos A)}}{\text{sin A(1 - cos A)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A - \text{cos A + cos}^2 A}{\text{sin A(1 - cos A)}}$

By formula,

sin² A + cos² A = 1

$\Rightarrow \dfrac{\text{1 - cos A}}{\text{sin A(1 - cos A)}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}} \\[1em] \Rightarrow \text{cosec A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin A}}{\text{1 - cos A}} - \text{cot A = cosec A}$ .

Question 2(v)

Prove the following identities :

$\sqrt{\dfrac{\text{1 - cos A}}{\text{1 + cos A}}} = \dfrac{\text{sin A}}{\text{1 + cos A}}$

Answer

Multiplying numerator and denominator of L.H.S. of above equation by $\sqrt{1 + \text{cos A}}$ :

$\Rightarrow \sqrt{\dfrac{1 - \text{cos A}}{\text{1 + cos A}}} \times \sqrt{\dfrac{1 + \text{cos A}}{1 + \text{cos A}}} \\[1em] \Rightarrow \sqrt{\dfrac{1 - \text{cos}^2 A}{(1 + \text{cos A})^2}}$

By formula,

1 - cos² A = sin² A

$\Rightarrow \dfrac{\sqrt{\text{sin}^2 A}}{1 + \text{cos A}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{1 + cos A}}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\dfrac{1 - \text{cos A}}{\text{1 + cos A}}} = \dfrac{\text{sin A}}{\text{1 + cos A}}$ .

Question 2(vi)

Prove the following identities :

$\dfrac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} = \text{2 tan A}$

Answer

By formula,

sec² - tan² A = 1

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} \\[1em] \Rightarrow \dfrac{\text{sec}^2 A - \text{tan}^2 A + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} \\[1em] \Rightarrow \dfrac{\text{(sec A - tan A)(sec A + tan A) + (sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} \\[1em] \Rightarrow \dfrac{\text{(sec A - tan A)[sec A + tan A + sec A - tan A]}}{\text{cosec A (sec A - tan A)}} \\[1em] \Rightarrow \dfrac{\text{2 sec A}}{\text{cosec A}} \\[1em] \Rightarrow \dfrac{2 \times \dfrac{1}{\text{cos A}}}{\dfrac{1}{\text{sin A}}} \\[1em] \Rightarrow 2\dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{2 tan A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{1 + \text{(sec A - tan A)}^2}{\text{cosec A (sec A - tan A)}} = \text{2 tan A}$ .

Question 2(vii)

Prove the following identities :

$\dfrac{\text{(cosec A - cot A)}^2 + 1}{\text{sec A (cosec A - cot A)}} = \text{2 cot A}$

Answer

By formula,

cosec² A - cot² A = 1

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{(cosec A - cot A)}^2 + 1}{\text{sec A (cosec A - cot A)}} \\[1em] \Rightarrow \dfrac{\text{(cosec A - cot A)}^2 + \text{cosec}^2 A - \text{cot}^2 A}{\text{sec A (cosec A - cot A)}} \\[1em] \Rightarrow \dfrac{\text{(cosec A - cot A)}^2 + \text{(cosec A - cot A)(cosec A + cot A)}}{\text{sec A (cosec A - cot A)}} \\[1em] \Rightarrow \dfrac{\text{(cosec A - cot A)(cosec A - cot A + cosec A + cot A)}}{\text{sec A (cosec A - cot A)}} \\[1em] \Rightarrow \dfrac{\text{2 cosec A}}{\text{sec A}} \\[1em] \Rightarrow \dfrac{\dfrac{2}{\text{sin A}}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{2 cos A}}{\text{sin A}} \\[1em] \Rightarrow \text{2 cot A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{(cosec A - cot A)}^2 + 1}{\text{sec A (cosec A - cot A)}} = \text{2 cot A}$ .

Question 2(viii)

Prove the following identities :

$\text{cot}^2 A \Big(\dfrac{\text{sec A - 1}}{\text{1 + sin A}}\Big) + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big)$ = 0

Answer

Solving L.H.S. of above equation :

$\Rightarrow \text{cot}^2 A \Big(\dfrac{\text{sec A - 1}}{\text{1 + sin A}}\times \dfrac{\text{sec A + 1}}{\text{sec A + 1}}\Big) + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \text{cot}^2 A \Big(\dfrac{\text{sec}^2 A - 1}{\text{(1 + sin A)(sec A + 1)}}\Big) + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big)$

By formula,

sec² A - 1 = tan² A

$\Rightarrow \dfrac{\text{cot}^2 A \text{ tan}^2 A}{\text{(1 + sin A)(sec A + 1)}} + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \dfrac{\text{cot}^2 A \times \dfrac{1}{\text{ cot}^2 A}}{\text{(1 + sin A)(sec A + 1)}} + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \dfrac{1}{\text{(1 + sin A)(sec A + 1)}} + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big) \\[1em] \Rightarrow \dfrac{1 + \text{sec}^2 A\text{(sin A - 1)(1 + sin A)}}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - \text{sec}^2 A\text{(1 - sin A)(1 + sin A)}}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - \text{sec}^2 A\text{(1 - sin}^2 A)}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - \text{sec}^2 A \times \text{cos}^2 A}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - \dfrac{1}{\text{cos}^2 A} \times \text{cos}^2 A}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{1 - 1}{\text{(1 + sin A)(sec A + 1)}} \\[1em] \Rightarrow 0.$

Since, L.H.S. = R.H.S.

Hence, proved that $\text{cot}^2 A \Big(\dfrac{\text{sec A - 1}}{\text{1 + sin A}}\Big) + \text{sec}^2 A \Big(\dfrac{\text{sin A - 1}}{\text{1 + sec A}}\Big)$ = 0.

Question 2(ix)

Prove the following identities :

$\dfrac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A}$ = 2 cos² A - 1

Answer

Solving L.H.S. of the above equation :

$\Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A} \\[1em] \Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{(\text{cos}^2 A - \text{sin}^2 A)(\text{cos}^2 A + \text{sin}^2 A)}$

By formula,

cos² A + sin² A = 1 and cos² A = 1 - sin² A.

$\Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{(1 - \text{sin}^2 A - \text{sin}^2 A)} \\[1em] \Rightarrow \dfrac{(1 - \text{2 sin}^2 A)^2}{(1 - \text{2 sin}^2 A)}\\[1em] \Rightarrow \text{1 - 2 sin}^2 A \\[1em] \Rightarrow 1 - 2(\text{1 - cos}^2 A) \\[1em] \Rightarrow 1 - 2 + \text{2 cos}^2 A \\[1em] \Rightarrow \text{2 cos}^2 A - 1.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{(1 - \text{2 sin}^2 A)^2}{\text{cos}^4 A - \text{sin}^4 A}$ = 2 cos² A - 1.

Question 2(x)

Prove the following identities :

sec⁴ A (1 - sin⁴ A) - 2 tan² A = 1

Answer

Solving L.H.S. of the above equation :

⇒ sec⁴ A (1 - sin⁴ A) - 2 tan² A

⇒ sec⁴ A (1 - sin² A)(1 + sin² A) - 2 tan² A

By formula,

1 - sin² A = cos² A

⇒ sec⁴ A cos² A (1 + sin² A) - 2 tan² A

⇒ sec⁴ A $\times \dfrac{1}{\text{sec}^2 A}$ (1 + sin² A) - 2 tan² A

⇒ sec² A (1 + sin² A) - 2 tan² A

⇒ sec² A + sec² A sin² A - 2 tan² A

⇒ sec² A + $\dfrac{1}{\text{cos}^2 A} \times$ sin² A - 2 tan² A

⇒ sec² A + tan² A - 2 tan² A

⇒ sec² A - tan² A

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that sec⁴ A (1 - sin⁴ A) - 2 tan² A = 1.

Question 2(xi)

Prove the following identities :

(1 + tan A + sec A)(1 + cot A - cosec A) = 2

Answer

Solving L.H.S. of the above equation :

$\Rightarrow \Big(1 + \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{1}{\text{cos A}}\Big)\Big(1 + \dfrac{\text{cos A}}{\text{sin A}} - \dfrac{1}{\text{sin A}}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{cos A + sin A + 1}}{\text{cos A}}\Big)\Big(\dfrac{\text{sin A + cos A - 1}}{\text{sin A}}\Big) \\[1em] \Rightarrow \dfrac{\text{(sin A + cos A)}^2 - (1)^2}{\text{sin A cos A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A + \text{2 sin A cos A} - 1}{\text{sin A cos A}}$

By formula,

sin² A + cos² A = 1

$\Rightarrow \dfrac{\text{1 - 1 + 2 sin A cos A}}{\text{sin A cos A}}\\[1em] \Rightarrow \dfrac{\text{2 sin A cos A}}{\text{sin A cos A}} \\[1em] \Rightarrow 2.$

Since, L.H.S. = R.H.S.

Hence, proved that (1 + tan A + sec A)(1 + cot A - cosec A) = 2.

Question 3

If x = a cos θ and y = b cot θ, show that :

$\dfrac{a^2}{x^2} - \dfrac{b^2}{y^2}$ = 1

Answer

Substituting value of x and y in L.H.S. of above equation :

$\Rightarrow \dfrac{a^2}{a^2 \text{ cos}^2 θ} - \dfrac{b^2}{b^2 \text{ cot}^2 θ} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 θ} - \dfrac{1}{\text{cot}^2 θ} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 θ} - \dfrac{1}{\dfrac{\text{cos}^2 θ}{\text{sin}^2 θ}} \\[1em] \Rightarrow \dfrac{1}{\text{cos}^2 θ} - \dfrac{\text{sin}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow \dfrac{1 - \text{sin}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow \dfrac{\text{cos}^2 θ}{\text{cos}^2 θ} \\[1em] \Rightarrow 1.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{a^2}{x^2} - \dfrac{b^2}{y^2}$ = 1.

Question 4

If sec A + tan A = p, show that :

sin A = $\dfrac{p^2 - 1}{p^2 + 1}$

Answer

Substituting value of p in R.H.S. of the above equation :

$\Rightarrow \dfrac{\text{(sec A + tan A)}^2 - 1}{\text{(sec A + tan A)}^2 + 1} \\[1em] \Rightarrow \dfrac{\text{sec}^2 A + \text{tan}^2 A + \text{2 sec A tan A} - 1}{\text{sec}^2 A + \text{tan}^2 A + \text{2 sec A tan A} + 1}$

By formula,

sec² A - 1 = tan² A and sec² A = 1 + tan² A

$\Rightarrow \dfrac{\text{sec}^2 A - 1 + \text{tan}^2 A + \text{2 sec A tan A}}{\text{sec}^2 A + \text{tan}^2 A + 1 + \text{2 sec A tan A}} \\[1em] \Rightarrow \dfrac{\text{tan}^2 A + \text{tan}^2 A + \text{2 sec A tan A}}{\text{sec}^2 A + \text{sec}^2 A + \text{2 sec A tan A}} \\[1em] \Rightarrow \dfrac{\text{2 tan}^2 A + \text{2 sec A tan A}}{\text{2 sec}^2 A + \text{2 sec A tan A}} \\[1em] \Rightarrow \dfrac{2\text{tan A}(\text{tan A + sec A})}{\text{2 sec A(sec A + tan A)}} \\[1em] \Rightarrow \dfrac{\text{tan A}}{\text{sec A}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{sin A}}{\text{cos A}}}{\dfrac{1}{\text{cos A}}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} \times \text{cos A} \\[1em] \Rightarrow \text{sin A}.$

Since, L.H.S. = R.H.S.

Hence, proved that sin A = $\dfrac{p^2 - 1}{p^2 + 1}$ .

Question 5(i)

If 2 sin A - 1 = 0, show that :

sin 3A = 3 sin A - 4 sin³ A

Answer

Given,

⇒ 2 sin A - 1 = 0

⇒ 2 sin A = 1

⇒ sin A = $\dfrac{1}{2}$

⇒ sin A = sin 30°

⇒ A = 30°.

Given equation : sin 3A = 3 sin A - 4 sin³ A

Substituting value of A in L.H.S. we get,

sin 3A = sin 3(30°) = sin 90° = 1.

Substituting value of A in R.H.S. we get,

⇒ 3 sin A - 4 sin³ A

⇒ 3 sin 30° - 4 sin³ 30°

⇒ $3 \times \dfrac{1}{2} - 4 \times \Big(\dfrac{1}{2}\Big)^3$

⇒ $\dfrac{3}{2} - 4 \times \dfrac{1}{8}$

⇒ $\dfrac{3}{2} - \dfrac{1}{2}$

⇒ $\dfrac{2}{2}$

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that sin 3A = 3 sin A - 4 sin³ A.

Question 5(ii)

If 4 cos² A - 3 = 0, show that :

cos 3A = 4 cos³ A - 3 cos A

Answer

Given,

⇒ 4 cos² A - 3 = 0

⇒ 4 cos² A = 3

⇒ cos² A = $\dfrac{3}{4}$

⇒ cos A = $\sqrt{\dfrac{3}{4}}$

⇒ cos A = $\dfrac{\sqrt{3}}{2}$ .

⇒ cos A = cos 30°

⇒ A = 30°.

L.H.S. = cos 3A = cos 3(30°) = cos 90° = 0.

R.H.S. = 4 cos³ A - 3 cos A

= 4 cos³ 30° - 3 cos 30°

= $4 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3 - 3 \times \dfrac{\sqrt{3}}{2}$

= $4 \times \dfrac{3\sqrt{3}}{8} - \dfrac{3\sqrt{3}}{2}$

= $\dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2}$

= 0.

Since, L.H.S. = R.H.S.

Hence, proved that cos 3A = 4 cos³ A - 3 cos A.

Question 6(i)

Evaluate :

$2\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big)$

Answer

Solving,

$\Rightarrow 2\Big(\dfrac{\text{tan 35°}}{\text{cot (90° - 35°)}}\Big)^2 + \Big(\dfrac{\text{cot (90° - 35°)}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{sec (90°- 50°)}}{\text{cosec 50°}}\Big)$

By formula,

cot(90° - θ) = tan θ and sec(90° - θ) = cosec θ

$\Rightarrow 2\Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big)^2 + \Big(\dfrac{\text{tan 35°}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{cosec 50°}}{\text{cosec 50°}}\Big) \\[1em] \Rightarrow 2 \times (1)^2 + (1)^2 - 3 \times 1 \\[1em] \Rightarrow 2 + 1 - 3 \\[1em] \Rightarrow 0.$

Hence, $2\Big(\dfrac{\text{tan 35°}}{\text{cot 55°}}\Big)^2 + \Big(\dfrac{\text{cot 55°}}{\text{tan 35°}}\Big)^2 - 3\Big(\dfrac{\text{sec 40°}}{\text{cosec 50°}}\Big)$ = 0.

Question 6(ii)

sec 26° sin 64° + $\dfrac{\text{cosec 33°}}{\text{sec 57°}}$

Answer

Solving,

⇒ sec 26° sin 64° + $\dfrac{\text{cosec 33°}}{\text{sec 57°}}$

⇒ sec (90° - 64°) sin 64° + $\dfrac{\text{cosec 33°}}{\text{sec (90° - 33°)}}$

By formula,

sec(90° - θ) = cosec θ

⇒ cosec 64° sin 64° + $\dfrac{\text{cosec 33°}}{\text{cosec 33°}}$

⇒ $\dfrac{1}{\text{sin 64°}} \times \text{sin 64°} + 1$

⇒ 1 + 1

⇒ 2.

Hence, sec 26° sin 64° + $\dfrac{\text{cosec 33°}}{\text{sec 57°}}$ = 2.

Question 6(iii)

$\dfrac{\text{5 sin 66°}}{\text{cos 24°}} - \dfrac{\text{2 cot 85°}}{\text{tan 5°}}$

Answer

Solving,

$\Rightarrow \dfrac{\text{5 sin 66°}}{\text{cos 24°}} - \dfrac{\text{2 cot 85°}}{\text{tan 5°}}$

By formula,

sin (90° - θ) = cos θ and cot (90° - θ) = tan θ

$\Rightarrow \dfrac{\text{5 sin (90° - 24°)}}{\text{cos 24°}} - \dfrac{\text{2 cot (90° - 5°)}}{\text{tan 5°}} \\[1em] \Rightarrow \dfrac{\text{5 cos 24°}}{\text{cos 24°}} - \dfrac{\text{2 tan 5°}}{\text{tan 5°}} \\[1em] \Rightarrow 5 - 2 \\[1em] \Rightarrow 3.$

Hence, $\dfrac{\text{5 sin 66°}}{\text{cos 24°}} - \dfrac{\text{2 cot 85°}}{\text{tan 5°}}$ = 3.

Question 6(iv)

3 cos 80° cosec 10° + 2 cos 59° cosec 31°

Answer

Given,

⇒ 3 cos (90° - 10°) cosec 10° + 2 cos 59° cosec (90° - 59°)

By formula,

cosec (90° - θ) = sec θ and cos(90° - θ) = sin θ

⇒ 3 sin 10° cosec 10° + 2 cos 59° sec 59°

⇒ 3 sin 10° $\times \dfrac{1}{\text{sin 10°}} + 2 \text{ cos 59°} \times \dfrac{1}{\text{cos 59°}}$

⇒ 3 + 2

⇒ 5.

Hence, 3 cos 80° cosec 10° + 2 cos 59° cosec 31° = 5.

Question 7(i)

Prove that :

tan (55° + x) = cot (35° - x)

Answer

Given equation : tan (55° + x) = cot (35° - x)

Solving L.H.S.

⇒ tan (55° + x)

⇒ tan [90° - (35° - x)]

By formula,

tan (90° - θ) = cot θ

⇒ cot (35° - x)

Since, L.H.S. = R.H.S.

Hence, proved that tan (55° + x) = cot (35° - x).

Question 7(ii)

sec (70° - θ) = cosec (20° + θ)

Answer

Given equation : sec (70° - θ) = cosec (20° + θ)

Solving L.H.S.

⇒ sec (70° - θ)

⇒ sec [90° - (20° + θ)]

By formula,

sec (90° - θ) = cosec θ

⇒ cosec (20° + θ)

Since, L.H.S. = R.H.S.

Hence, proved that sec (70° - θ) = cosec (20° + θ).

Question 7(iii)

sin(28° + A) = cos(62° - A)

Answer

Given equation : sin(28° + A) = cos(62° - A)

Solving L.H.S.

⇒ sin(28° + A)

⇒ sin [90° - (62° - A)]

By formula,

sin (90° - θ) = cos θ

⇒ cos (62° - A)

Since, L.H.S. = R.H.S.

Hence, proved that sin(28° + A) = cos(62° - A).

Question 7(iv)

$\dfrac{1}{\text{1 + cos (90° - A)}} + \dfrac{1}{\text{1 - cos (90° - A)}}$ = 2 cosec² (90° - A)

Answer

By formula,

cos (90° - A) = sin A

Solving L.H.S. of above equation,

$\Rightarrow \dfrac{1}{\text{1 + sin A}} + \dfrac{1}{1 - \text{sin A}} \\[1em] \Rightarrow \dfrac{\text{1 - sin A + 1 + sin A}}{\text{(1 + sin A)(1 - sin A)}} \\[1em] \Rightarrow \dfrac{2}{1 - \text{sin}^2 A} \\[1em] \Rightarrow \dfrac{2}{\text{cos}^2 A} \\[1em] \Rightarrow 2\text{ sec}^2 A \\[1em] \Rightarrow 2\text{ cosec}^2 (90° - A) \space [\because \text{ sec A} = \text{ cosec(90° - A)}]$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{1}{\text{1 + cos (90° - A)}} + \dfrac{1}{\text{1 - cos (90° - A)}}$ = 2 cosec² (90° - A).

Question 8

If A and B are complementary angles, prove that :

(i) cot B + cos B = sec A cos B (1 + sin B)

(ii) cot A cot B - sin A cos B - cos A sin B = 0

(iii) cosec² A + cosec² B = cosec² A cosec² B

(iv) $\dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos B - cos A}}{\text{cos B + cos A}} = \dfrac{2}{\text{2 sin}^2 A - 1}$

Answer

Given,

A + B = 90°

B = 90° - A and A = 90° - B.

(i) Substituting value of B in L.H.S. of equation :

$\Rightarrow \text{cot (90° - A) + cos (90° - A)} \\[1em] \Rightarrow \text{tan A + sin A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + \text{sin A} \\[1em] \Rightarrow \dfrac{\text{sin A + sin A cos A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{sin A(1 + cos A)}}{\text{cos A}}\\[1em] \Rightarrow \text{sin A sec A (1 + cos A)} \\[1em] \Rightarrow \text{sin (90° - B) sec A [1 + cos (90° - B)]}$

By formula,

sin (90° - B) = cos B and cos (90° - B) = sin B

⇒ cos B sec A (1 + sin B).

Since, L.H.S. = R.H.S.

Hence, proved that cot B + cos B = sec A cos B (1 + sin B).

(ii) Substituting value of B in L.H.S. of equation :

⇒ cot A cot (90° - A) - sin A cos (90° - A) - cos A sin (90° - A)

By formula,

cot (90° - A) = tan A, cos (90° - A) = sin A and sin (90° - A) = cos A

⇒ cot A tan A - sin A sin A - cos A cos A

⇒ cot A $\times \dfrac{1}{\text{cot A}}$ - (sin² A + cos² A)

⇒ 1 - 1

⇒ 0.

Since, L.H.S. = R.H.S.

Hence, proved that cot A cot B - sin A cos B - cos A sin B = 0.

(iii) Substituting value of B in L.H.S. of equation :

⇒ cosec² A + cosec² (90° - A)

By formula,

cosec (90° - A) = sec A

⇒ $\dfrac{1}{\text{sin}^2 A}$ + sec² A

⇒ $\dfrac{1}{\text{sin}^2 A} + \dfrac{1}{\text{cos}^2 A}$

⇒ $\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{sin}^2 A \text{cos}^2 A}$

By formula,

sin² A + cos² A = 1 and sec (90° - A) = cosec A

⇒ $\dfrac{1}{\text{sin}^2 A \text{cos}^2 A}$

⇒ cosec² A sec² A

⇒ cosec² A sec² (90° - B)

⇒ cosec² A cosec² B

Since, L.H.S. = R.H.S.

Hence, proved that cosec² A + cosec² B = cosec² A cosec² B.

(iv) Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos B - cos A}}{\text{cos B + cos A}} \\[1em] \Rightarrow \dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos (90° - A) - cos (90° - B)}}{\text{cos (90° - A) + cos (90° - B)}}$

By formula,

cos (90° - θ) = sin θ and sin (90° - θ) = cos θ.

$\Rightarrow \dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{sin A - sin B}}{\text{sin A + sin B}} \\[1em] \Rightarrow \dfrac{\text{(sin A + sin B)}^2 + \text{(sin A - sin B)}^2}{\text{(sin A - sin B)(sin A + sin B)}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{sin}^2 B + \text{2 sin A sin B + sin}^2 A + \text{sin}^2 B - \text{2 sin A sin B}}{\text{sin}^2 A - \text{sin}^2 B} \\[1em] \Rightarrow 2 \dfrac{\text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A - \text{sin}^2 B} \\[1em] \Rightarrow 2 \dfrac{\text{sin}^2 A + \text{sin}^2 (90° - A)}{\text{sin}^2 A - \text{sin}^2 (90° - A)} \\[1em] \Rightarrow 2 \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \dfrac{2 \times 1}{\text{sin}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \dfrac{2}{\text{sin}^2 A - \text{cos}^2 A}$

By formula,

cos² A = 1 - sin² A

$\Rightarrow \dfrac{2}{\text{sin}^2 A - (1 - \text{sin}^2 A)} \\[1em] \Rightarrow \dfrac{2}{\text{sin}^2 A + \text{sin}^2 A - 1} \\[1em] \Rightarrow \dfrac{2}{\text{2 sin}^2 A - 1}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{sin A + sin B}}{\text{sin A - sin B}} + \dfrac{\text{cos B - cos A}}{\text{cos B + cos A}} = \dfrac{2}{\text{2 sin}^2 A - 1}$ .

Question 9(i)

Prove that :

$\dfrac{1}{\text{sin A - cos A}} - \dfrac{1}{\text{sin A + cos A}} = \dfrac{\text{2 cos A}}{\text{2 sin}^2 A - 1}$

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1}{\text{sin A - cos A}} - \dfrac{1}{\text{sin A + cos A}} \\[1em] \Rightarrow \dfrac{\text{sin A + cos A - (sin A - cos A)}}{\text{(sin A - cos A)(sin A + cos A)}} \\[1em] \Rightarrow \dfrac{\text{sin A - sin A + cos A + cos A}}{\text{sin}^2 A - \text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 cos A}}{\text{sin}^2 A - \text{cos}^2 A}$

By formula,

cos² A = 1 - sin² A

$\Rightarrow \dfrac{\text{2 cos A}}{\text{sin}^2 A - (1 - \text{sin}^2 A)} \\[1em] \Rightarrow \dfrac{\text{2 cos A}}{\text{sin}^2 A + \text{sin}^2 A - 1} \\[1em] \Rightarrow \dfrac{\text{2 cos A}}{\text{2 sin}^2 A - 1}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{1}{\text{sin A - cos A}} - \dfrac{1}{\text{sin A + cos A}} = \dfrac{\text{2 cos A}}{\text{2 sin}^2 A - 1}$ .

Question 9(ii)

Prove that :

$\dfrac{\text{cot}^2 A}{\text{cosec A - 1}} - 1$ = cosec A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{cot}^2 A}{\text{cosec A - 1}} - 1 \\[1em] \Rightarrow \dfrac{\text{cot}^2 A - \text{(cosec A - 1)}}{\text{cosec A - 1}} \\[1em] \Rightarrow \dfrac{\text{cot}^2 A + 1 - \text{cosec A}}{\text{cosec A - 1}}$

By formula,

cot² A + 1 = cosec² A

$\Rightarrow \dfrac{\text{cosec}^2 A - \text{cosec A}}{\text{cosec A - 1}} \\[1em] \Rightarrow \dfrac{\text{cosec A(cosec A - 1)}}{\text{cosec A - 1}} \\[1em] \Rightarrow \text{cosec A}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cot}^2 A}{\text{cosec A - 1}} - 1$ = cosec A.

Question 9(iii)

Prove that :

$\dfrac{\text{cos A}}{\text{1 + sin A}}$ = sec A - tan A

Answer

Solving R.H.S. of the above equation :

$\Rightarrow \text{sec A - tan A} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{1 - sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{1 - sin A}}{\text{cos A}} \times \dfrac{\text{1 + sin A}}{\text{1 + sin A}} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 A}{\text{cos A(1+ sin A)}} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{cos A(1 + sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos A}}{\text{1 + sin A}}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cos A}}{\text{1 + sin A}}$ = sec A - tan A.

Question 9(iv)

Prove that :

cos A(1 + cot A) + sin A(1 + tan A) = sec A + cosec A

Answer

Solving L.H.S. of the above equation :

⇒ cos A(1 + cot A) + sin A(1 + tan A)

⇒ cos A + cos A cot A + sin A + sin A tan A

⇒ cos A + cos A $\times \dfrac{\text{cos A}}{\text{sin A}} + \text{sin A} + \text{sin A} \times \dfrac{\text{sin A}}{\text{cos A}}$

⇒ cos A + $\dfrac{\text{sin}^2 A}{\text{cos A}} + \text{ sin A } + \dfrac{\text{cos}^2 A}{\text{sin A}}$

⇒ $\dfrac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos A}} + \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A}}$

By formula,

sin² A + cos² A = 1.

⇒ $\dfrac{1}{\text{cos A}} + \dfrac{1}{\text{sin A}}$

⇒ sec A + cosec A.

Since, L.H.S. = R.H.S.

Hence, proved that cos A(1 + cot A) + sin A(1 + tan A) = sec A + cosec A.

Question 9(v)

$\sqrt{\text{sec}^2 A + \text{cosec}^2 A}$ = tan A + cot A

Answer

Solving L.H.S. of the above equation :

$\Rightarrow \sqrt{\Big(\dfrac{1}{\text{cos A}}\Big)^2 + \Big(\dfrac{1}{\text{sin A}}\Big)^2} \\[1em] \Rightarrow \sqrt{\dfrac{1}{\text{cos}^2 A} + \dfrac{1}{\text{sin}^2 A}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos}^2 A\text{ sin}^2 A}}$

By formula,

sin² A + cos² A = 1

$\Rightarrow \sqrt{\dfrac{1}{\text{cos}^2 A\text{ sin}^2 A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A}\text{ cos A}}.$

Solving R.H.S. of the equation :

$\Rightarrow \text{tan A + cot A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{cos A sin A}} \\[1em] \Rightarrow \dfrac{1}{\text{sin A cos A}}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\sqrt{\text{sec}^2 A + \text{cosec}^2 A}$ = tan A + cot A.

Question 9(vi)

(sin A + cos A)(sec A + cosec A) = 2 + sec A cosec A

Answer

Solving L.H.S. of the above equation :

$\Rightarrow \text{sin A sec A + sin A cosec A + cos A sec A + cos A cosec A} \\[1em] \Rightarrow \text{sin A} \times \dfrac{1}{\text{cos A}} + \text{sin A} \times \dfrac{1}{\text{sin A}} + \text{cos A} \times \dfrac{1}{\text{cos A}} + \text{cos A} \times \dfrac{1}{\text{sin A}} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} + 1 + 1 + \dfrac{\text{cos A}}{\text{sin A}} \\[1em] \Rightarrow 2 + \dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}} \\[1em] \Rightarrow 2 + \dfrac{1}{\text{sin A cos A}} \\[1em] \Rightarrow 2 + \text{cosec A sec A}.$

Since, L.H.S. = R.H.S.

Hence, proved that (sin A + cos A)(sec A + cosec A) = 2 + sec A cosec A.

Question 9(vii)

(tan A + cot A)(cosec A - sin A)(sec A - cos A) = 1

Answer

Solving L.H.S. of the above equation :

$\Rightarrow \text{(tan A + cot A)(cosec A - sin A)(sec A - cos A)} \\[1em] \Rightarrow \Big(\dfrac{\text{sin A}}{\text{cos A}} + \dfrac{\text{cos A}}{\text{sin A}}\Big)\Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big)\Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin A cos A}}\Big)\Big(\dfrac{1 - \text{sin}^2 A}{\text{sin A}}\Big)\Big(\dfrac{1 - \text{cos}^2 A}{\text{cos A}}\Big)$

By formula,

sin² A + cos² A = 1, 1 - sin² A = cos² A and 1 - cos² A = sin² A.

$\Rightarrow \dfrac{1}{\text{sin A cos A}} \times \dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A \text{cos}^2 A}{\text{sin}^2 A \text{cos}^2 A} \\[1em] \Rightarrow 1.$

Since, L.H.S. = R.H.S.

Hence, proved that (tan A + cot A)(cosec A - sin A)(sec A - cos A) = 1.

Question 9(viii)

cot² A - cot² B = $\dfrac{\text{cos}^2 \text{A} - \text{cos}^2 \text{B}}{\text{sin}^2 \text{A} \text{ sin}^2 \text{B}}$ = cosec² A - cosec² B

Answer

Solving,

$\Rightarrow \text{cot}^2 A - \text{cot}^2 B \\[1em] \Rightarrow \dfrac{\text{cos}^2 A}{\text{sin}^2 A} - \dfrac{\text{cos}^2 B}{\text{sin}^2 B} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A \text{ sin}^2 B - \text{cos}^2 B \text{ sin}^2 A}{\text{sin}^2 A \text{ sin}^2 B} \\[1em]$

By formula,

sin² θ = 1 - cos² θ

$\Rightarrow \dfrac{\text{cos}^2 A (1 - \text{ cos}^2 B) - \text{cos}^2 B (1 - \text{ cos}^2 A)}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - \text{cos}^2 A \text{cos}^2 B - \text{cos}^2 B + \text{cos}^2 A \text{cos}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{\text{cos}^2 A - \text{cos}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 A - (1 - \text{sin}^2 B)}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{1 - 1 - \text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow \dfrac{-\text{sin}^2 A + \text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow -\dfrac{\text{sin}^2 A}{\text{sin}^2 A \text{ sin}^2 B} + \dfrac{\text{sin}^2 B}{\text{sin}^2 A \text{ sin}^2 B} \\[1em] \Rightarrow -\dfrac{1}{\text{sin}^2 B} + \dfrac{1}{\text{sin}^2 A} \\[1em] \Rightarrow -\text{cosec}^2 B + \text{cosec}^2 A \\[1em] \Rightarrow \text{cosec}^2 A - \text{cosec}^2 B.$

Hence, proved that cot² A - cot² B = $\dfrac{\text{cos}^2 \text{A} - \text{cos}^2 \text{B}}{\text{sin}^2 \text{A} \text{ sin}^2 \text{B}}$ = cosec² A - cosec² B.

Question 9(ix)

Prove that :

$\dfrac{\text{cot A - 1}}{\text{2 - sec}^2 A} = \dfrac{\text{cot A}}{\text{1 + tan A}}$

Answer

Solving L.H.S. of the above equation :

$\Rightarrow \dfrac{\dfrac{1}{\text{tan A}} - 1}{\text{2 - (1 + tan}^2 A)} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - tan A}}{\text{tan A}}}{2 - 1 - \text{tan}^2 A} \\[1em] \Rightarrow \dfrac{\text{1 - tan A}}{\text{tan A(1 - tan}^2 A)} \\[1em] \Rightarrow \dfrac{\text{1 - tan A}}{\text{tan A(1 - tan A)(1 + tan A)}} \\[1em] \Rightarrow \dfrac{1}{\text{tan A(1 + tan A)}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{1}{\text{cot A}} \text{(1 + tan A)}} \\[1em] \Rightarrow \dfrac{\text{cot A}}{\text{1 + tan A}}.$

Since, L.H.S. = R.H.S.

Hence, proved that $\dfrac{\text{cot A - 1}}{\text{2 - sec}^2 A} = \dfrac{\text{cot A}}{\text{1 + tan A}}$ .

Question 10

If 4 cos² A - 3 = 0 and 0° ≤ A ≤ 90°; then prove that :

(i) sin 3A = 3 sin A - 4 sin³ A

(ii) cos 3A = 4 cos³ A - 3 cos A

Answer

Given,

⇒ 4 cos² A - 3 = 0

⇒ 4 cos² A = 3

⇒ cos² A = $\dfrac{3}{4}$

⇒ cos A = $\sqrt{\dfrac{3}{4}} = \dfrac{\sqrt{3}}{2}$ .

⇒ cos A = cos 30°

⇒ A = 30°.

(i) To prove:

sin 3A = 3 sin A - 4 sin³ A

Solving L.H.S. of the equation :

⇒ sin 3A = sin 3(30°)

= sin 90° = 1.

Solving R.H.S. of the equation :

⇒ 3 sin A - 4 sin³ A

⇒ 3 sin 30° - 4 sin³ 30°

⇒ 3 $\times \dfrac{1}{2} - 4 \times \Big(\dfrac{1}{2}\Big)^3$

⇒ $\dfrac{3}{2} - 4 \times \dfrac{1}{8}$

⇒ $\dfrac{3}{2} - \dfrac{1}{2}$

⇒ $\dfrac{2}{2}$

⇒ 1.

Since, L.H.S. = R.H.S.

Hence, proved that sin 3A = 3 sin A - 4 sin³ A.

(ii) To prove:

cos 3A = 4 cos³ A - 3 cos A

Solving L.H.S.

⇒ cos 3A = cos 3(30°) = cos 90° = 0.

Solving R.H.S.

⇒ 4 cos³ A - 3 cos A

⇒ 4 cos³ 30° - 3 cos 30°

⇒ 4 $\times \Big(\dfrac{\sqrt{3}}{2}\Big)^3 - 3 \times \dfrac{\sqrt{3}}{2}$

⇒ 4 $\times \dfrac{3\sqrt{3}}{8} - \dfrac{3\sqrt{3}}{2}$

⇒ $\dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2}$

⇒ 0.

Since, L.H.S. = R.H.S.

Hence, proved that cos 3A = 4 cos³ A - 3 cos A

Question 11

Find A, if 0° ≤ A ≤ 90° and :

(i) 2 cos² A - 1 = 0

(ii) sin 3A - 1 = 0

(iii) 4 sin² A - 3 = 0

(iv) cos² A - cos A = 0

(v) 2 cos² A + cos A - 1 = 0

Answer

(i) Solving,

⇒ 2 cos² A - 1 = 0

⇒ 2 cos² A = 1

⇒ cos² A = $\dfrac{1}{2}$

⇒ cos A = $\sqrt{\dfrac{1}{2}}$

⇒ cos A = $\dfrac{1}{\sqrt{2}}$

⇒ cos A = cos 45°

⇒ A = 45°.

Hence, A = 45°.

(ii) Solving,

⇒ sin 3A - 1 = 0

⇒ sin 3A = 1

⇒ sin 3A = sin 90°

⇒ 3A = 90°

⇒ A = 30°.

Hence, A = 30°.

(iii) Solving,

⇒ 4 sin² A - 3 = 0

⇒ 4 sin² A = 3

⇒ sin² A = $\dfrac{3}{4}$

⇒ sin A = $\sqrt{\dfrac{3}{4}}$

⇒ sin A = $\dfrac{\sqrt{3}}{2}$

⇒ sin A = sin 60°

⇒ A = 60°.

Hence, A = 60°.

(iv) Solving,

⇒ cos² A - cos A = 0

⇒ cos A(cos A - 1) = 0

⇒ cos A = 0 or cos A - 1 = 0

⇒ cos A = 0 or cos A = 1

⇒ cos A = cos 90° or cos A = cos 0°

⇒ A = 90° or A = 0°.

Hence, A = 0° or 90°.

(v) Solving,

⇒ 2 cos² A + cos A - 1 = 0

⇒ 2 cos² A + 2 cos A - cos A - 1 = 0

⇒ 2 cos A(cos A + 1) - 1(cos A + 1) = 0

⇒ (2 cos A - 1)(cos A + 1) = 0

⇒ 2 cos A = 1 or cos A = -1

⇒ cos A = $\dfrac{1}{2}$ or cos A = -1

Since, cos A cannot be negative in the range 0° ≤ A ≤ 90°.

∴ cos A = $\dfrac{1}{2}$

⇒ cos A = cos 60°

⇒ A = 60°

Hence, A = 60°.

Question 12

If 0° < A < 90°; find A if :

(i) $\dfrac{\text{cos A}}{\text{1 - sin A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = 4$

(ii) $\dfrac{\text{sin A}}{\text{sec A - 1}} + \dfrac{\text{sin A}}{\text{sec A + 1}}$ = 2

Answer

(i) Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{cos A}}{\text{1 - sin A}} + \dfrac{\text{cos A}}{\text{1 + sin A}} = 4 \\[1em] \Rightarrow \dfrac{\text{cos A(1 + sin A) + cos A(1 - sin A)}}{\text{(1 + sin A)(1 - sin A)}} \\[1em] \Rightarrow \dfrac{\text{cos A + cos A sin A + cos A - cos A sin A}}{\text{1 - sin}^2 A}$

By formula,

1 - sin² A = cos² A

$\Rightarrow \dfrac{\text{2 cos A}}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{2}{\text{cos A}} \\[1em] \Rightarrow 2\text{ sec A}.$

Given, R.H.S. = 4

∴ 2 sec A = 4

⇒ sec A = 2

⇒ sec A = sec 60°

⇒ A = 60°.

Hence, A = 60°.

(ii) Solving L.H.S. of the equation :

$\Rightarrow \dfrac{\text{sin A}}{\text{sec A - 1}} + \dfrac{\text{sin A}}{\text{sec A + 1}} = 2 \\[1em] \Rightarrow \dfrac{\text{sin A(sec A + 1) + sin A(sec A - 1)}}{\text{(sec A - 1)(sec A + 1)}} \\[1em] \Rightarrow \dfrac{\text{sin A sec A + sin A + sin A sec A - sin A}}{\text{sec}^2 A - 1} \\[1em] \Rightarrow \dfrac{\text{2 sin A sec A}}{\text{sec}^2 A - 1}$

By formula,

sec² A - 1 = tan² A

$\Rightarrow \dfrac{\text{2 sin A} \times \dfrac{1}{\text{cos A}}}{\text{tan}^2 A} \\[1em] \Rightarrow \dfrac{\text{2 tan A}}{\text{tan}^2 A} \\[1em] \Rightarrow \dfrac{2}{\text{tan A}} \\[1em] \Rightarrow \text{2 cot A}.$

Given, R.H.S. = 2

∴ 2 cot A = 2

⇒ cot A = 1

⇒ cot A = cot 45°

⇒ A = 45°.

Hence, A = 45°.

Question 13

Prove that :

(cosec A - sin A)(sec A - cos A) sec² A = tan A

Answer

Solving L.H.S. of the equation :

$\Rightarrow \Big(\dfrac{1}{\text{sin A}} - \text{sin A}\Big) \Big(\dfrac{1}{\text{cos A}} - \text{cos A}\Big)\text{sec}^2 \text{A} \\[1em] \Rightarrow \Big(\dfrac{\text{1 - sin}^2 A}{\text{sin A}}\Big) \times \Big(\dfrac{\text{1 - cos}^2 A}{\text{cos A}}\Big) \times \text{sec}^2 A$

By formula,

1 - sin² A = cos² A and 1 - cos² A = sin² A

$\Rightarrow \dfrac{\text{cos}^2 A}{\text{sin A}} \times \dfrac{\text{sin}^2 A}{\text{cos A}} \times \dfrac{1}{\text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{tan A}.$

Since, L.H.S. = R.H.S.

Hence, proved that (cosec A - sin A)(sec A - cos A) sec² A = tan A

Question 14

Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.

Answer

Given equation,

⇒ (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.

Solving L.H.S. of the equation :

$\Rightarrow \text{(sin θ + cos θ)(tan θ + cot θ)} \\[1em] \Rightarrow \text{(sin θ + cos θ)}\Big(\dfrac{\text{sin θ}}{\text{cos θ}} + \dfrac{\text{cos θ}}{\text{sin θ}}\Big) \\[1em] \Rightarrow \text{(sin θ + cos θ)}\Big(\dfrac{\text{sin}^2 θ + \text{cos}^2 θ}{\text{cos θ sin θ}}\Big)$

By formula,

sin² θ + cos² θ = 1.

$\Rightarrow \text{(sin θ + cos θ)} \times \dfrac{1}{\text{cos θ sin θ}} \\[1em] \Rightarrow \dfrac{\text{sin θ}}{\text{cos θ sin θ}} + \dfrac{\text{cos θ}}{\text{cos θ sin θ}} \\[1em] \Rightarrow \dfrac{1}{\text{cos θ}} + \dfrac{1}{\text{sin θ}} \\[1em] \Rightarrow \text{sec θ + cosec θ}.$

Hence, proved that (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.

Question 15

Evaluate without using trigonometric tables,

sin² 28° + sin² 62° + tan² 38° - cot² 52° + $\dfrac{1}{4} \text{sec}^2 \space 30°$

Answer

Solving,

⇒ sin² 28° + sin² 62° + tan² 38° - cot² 52° + $\dfrac{1}{4} \text{sec}^2 \space 30°$

⇒ sin² 28° + sin² (90° - 28°) + tan² (90° - 52°) - cot² 52° + $\dfrac{1}{4} \times \Big(\dfrac{2}{\sqrt{3}}\Big)^2$

By formula,

sin(90° - θ) = cos θ and tan(90° - θ) = cot θ

⇒ sin² 28° + cos² 28° + cot² 52° - cot² 52° + $\dfrac{1}{4} \times \dfrac{4}{3}$

By formula,

sin² θ + cos² θ = 1

⇒ 1 + $\dfrac{1}{3}$

⇒ $1\dfrac{1}{3}$ .

Hence, sin² 28° + sin² 62° + tan² 38° - cot² 52° + $\dfrac{1}{4} \text{sec}^2 30° = 1\dfrac{1}{3}$ .

Chapter 21

Trigonometrical Identities

Class - 10 Concise Mathematics Selina

Exercise 21(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Exercise 21(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2(i)

Question 2(ii)

Question 2(iii)

Question 2(iv)

Question 2(v)

Question 2(vi)

Question 2(vii)

Question 2(viii)

Question 2(ix)

Question 3

Question 4

Question 5

Question 6

Exercise 21(C)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2(i)

Question 2(ii)

Question 3

Question 4(i)

Question 4(ii)

Question 5

Question 6(i)

Question 6(ii)

Question 6(iii)

Question 6(iv)

Question 6(v)

Question 6(vi)

Question 6(vii)

Question 7