Trigonometrical Identities — Exercise 21(A)

sin⁴ θ - cos⁴ θ is equal to :

1. sin² θ - 1

2. 1 + cos² θ

3. 2 sin² θ - 1

4. 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.

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

1. 2 cosec² θ

2. 2 sec² θ

3. cosec² θ

4. 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.

cot⁴ θ + cot² θ is equal to :

1. 2 cot² θ. cosec² θ

2. tan² θ + tan⁴ θ

3. tan² θ.cosec² θ

4. 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.

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

1. 1 + sin A

2. sec² A (1 + sin A)

3. sec² A

4. 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.

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

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

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

3. $\dfrac{1}{\text{1 + cos A}}$

4. $\dfrac{1}{\text{1 - cos A}}$

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.

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

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

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.

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.

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.

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

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.

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.

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.

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.

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.

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.

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

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.

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.

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

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

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.

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.

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.

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.

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.

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.

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.

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

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

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).

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.