Trigonometrical Identities — Test Yourself

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

1. cot γ < cot β

2. cot γ > cot α

3. cot β > cot α

4. 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.

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

1. 1

2. 3

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

4. $\dfrac{1}{3}$

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.

(1 + tan² A) × cos² A - 1 is :

1. 1

2. -2

3. 0

4. 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.

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

1. cos A

2. sec A

3. cosec A

4. 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.

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 :

1. A and B

2. B and C

3. C and D

4. 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.

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

1. x² - y²

2. y² - x²

3. x² + y²

4. (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.

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

1. tan 30°

2. tan 45°

3. tan 60°

4. 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.

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

1. 0

2. -1

3. 1

4. 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.

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

1. 45°

2. 90°

3. 30°

4. 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.

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.

1. A is true, R is false.

2. A is false, R is true.

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

4. 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.

(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

1. A is true, R is false.

2. A is false, R is true.

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

4. 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.

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°

1. A is true, R is false.

2. A is false, R is true.

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

4. 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.

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

Assertion (A): x = sin A

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

1. A is true, R is false.

2. A is false, R is true.

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

4. 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.

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

1. Both the statements are true.

2. Both the statements are false.

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

4. 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.

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

Statement (1): x = 1

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

1. Both the statement are true.

2. Both the statement are false.

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

4. 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.

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

1. Both the statement are true.

2. Both the statement are false.

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

4. 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.

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°

1. Both the statement are true.

2. Both the statement are false.

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

4. 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.

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

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

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

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

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

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

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

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.

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.

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.

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.

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.

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

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).

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).

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).

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

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

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

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.

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.

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.

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

(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.

(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.

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.

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

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

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°.

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°.

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

Prove that :

(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.

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° + $\dfrac{1}{4} \text{sec}^2 30° = 1\dfrac{1}{3}$.

Prove that :

(i) (1 + cot² A)(1 - cos A)(1 + cos A) = 1

(ii) $\dfrac{\text{sin}^4 A + \text{cos}^4 A}{\text{sin}^2 A. \text{cos}^2 A}$ = tan² A + cot² A

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

(iv) $\dfrac{\text{cos}^2 A}{\text{1 - tan A}} + \dfrac{\text{sin}^3 A}{\text{sin A - cos A}}$ = 1 + sin A cos A

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

Answer

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

⇒ (1 + cot² A)(1 - cos A)(1 + cos A)

⇒ (1 + cot² A)(1 - cos² A)

By formula,

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

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

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

Hence, proved that (1 + cot² A)(1 - cos A)(1 + cos A) = 1.

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

$\Rightarrow \dfrac{\text{sin}^4 A + \text{cos}^4 A}{\text{sin}^2 A. \text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^4 A}{\text{sin}^2 A. \text{cos}^2 A} + \dfrac{\text{cos}^4 A}{\text{sin}^2 A. \text{cos}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A}{\text{cos}^2 A} + \dfrac{\text{cos}^2 A}{\text{sin}^2 A} \\[1em] \Rightarrow \text{tan}^2 A + \text{cot}^2 A.$

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

Hence, proved that $\dfrac{\text{sin}^4 A + \text{cos}^4 A}{\text{sin}^2 A. \text{cos}^2 A}$ = tan² A + cot² A.

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

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

By formula,

1 - cos² A = sin² A.

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

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

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

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

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

By formula,

a³ - b³ = (a - b)(a² + ab + b²)

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

By formula,

sin² A + cos² A = 1.

⇒ 1 + sin A cos A.

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

Hence, proved that $\dfrac{\text{cos}^2 A}{\text{1 - tan A}} + \dfrac{\text{sin}^3 A}{\text{sin A - cos A}}$ = 1 + sin A cos A.

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

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

Multiplying numerator and denominator by (1 + sin A + cos A) :

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

By formula,

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

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