Trigonometrical Ratios — Exercise 21(E)

If 2sin A - 1 = 0 and A is an acute angle, the measure of angle A is :

1. 30°

2. 45°

3. 60°

4. 90°

Answer

2sin A - 1 = 0

⇒ 2sin A = 1

⇒ sin A = $\dfrac{1}{2}$

⇒ sin A = sin 30°

⇒ A = 30°

Hence, option 1 is the correct option.

If cos A (cos A - 1) = 0; the measure of angle A is :

1. 90° or 0°

2. 90° and 0°

3. 45° or 90°

4. 45° and 90°

Answer

cos A (cos A - 1) = 0

⇒ cos A = 0 or cos A = 1

⇒ cos A = cos 90° or cos A = cos 0°

The measure of angle A is 90° or 0°.

Hence, option 1 is the correct option.

If tan⁴ A - 1 = 0 and angle A is acute, then A is :

1. 30°

2. 45°

3. ± 45°

4. ± 30°

Answer

tan⁴ A - 1 = 0

⇒ tan⁴A = 1

⇒ (tan A)⁴ = 1

⇒ tan A = tan 45°

⇒ A = 45°

Hence, option 2 is the correct option.

If 2sin 3A - 1 = 0 ; the value of angle A is :

1. 20°

2. 60°

3. 10°

4. 30°

Answer

2sin 3A - 1 = 0

⇒ 2sin 3A = 1

⇒ sin 3A = $\dfrac{1}{2}$

⇒ sin 3A = sin 30°

So, 3A = 30°

⇒ A = $\dfrac{30°}{3}$

⇒ A = 10°

Hence, option 3 is the correct option.

If 3tan² A - 1 = 0 and angle A is acute, the measure of angle A is :

1. 20°

2. 45°

3. 60°

4. 30°

Answer

3tan² A - 1 = 0

⇒ 3tan² A = 1

⇒ tan² A = $\dfrac{1}{3}$

⇒ tan A = $\sqrt{\dfrac{1}{3}}$

⇒ tan A = $\dfrac{1}{\sqrt3}$

⇒ tan A = tan 30°

⇒ A = 30°

Hence, option 4 is the correct option.

Solve the following equations for A, if :

2 sin A = 1

Answer

2 sin A = 1

⇒ sin A = $\dfrac{1}{2}$

⇒ sin A = sin 30°

Hence, A = 30°.

Solve the following equations for A, if :

2 cos 2 A = 1

Answer

2 cos 2 A = 1

⇒ cos 2A = $\dfrac{1}{2}$

⇒ cos 2A = cos 60°

So, 2A = 60°

⇒ A = $\dfrac{60°}{2}$

⇒ A = 30°

Hence, A = 30°.

Solve the following equations for A, if :

sin 3 A = $\dfrac{\sqrt3}{2}$

Answer

sin 3 A = $\dfrac{\sqrt3}{2}$

⇒ sin 3A = sin 60°

So, 3A = 60°

⇒ A = $\dfrac{60°}{3}$

⇒ A = 20°

Hence, A = 20°.

Solve the following equations for A, if :

sec 2 A = 2

Answer

sec 2 A = 2

⇒ sec 2A = sec 60°

So, 2A = 60°

⇒ A = $\dfrac{60°}{2}$

⇒ A = 30°

Hence, A = 30°.

Solve the following equations for A, if :

${\sqrt3}$ tan A = 1

Answer

${\sqrt3}$ tan A = 1

⇒ tan A = $\dfrac{1}{\sqrt3}$

⇒ tan A = tan 30°

So, A = 30°

Hence, A = 30°.

Solve the following equations for A, if :

tan 3 A = 1

Answer

tan 3 A = 1

⇒ tan 3A = tan 45°

So, 3A = 45°

⇒ A = $\dfrac{45°}{3}$

⇒ A = 15°

Hence, A = 15°.

Solve the following equations for A, if :

2 sin 3 A = 1

Answer

2 sin 3 A = 1

⇒ sin 3A = $\dfrac{1}{2}$

⇒ sin 3A = sin 30°

So, 3A = 30°

⇒ A = $\dfrac{30°}{3}$

⇒ A = 10°

Hence, A = 10°.

Solve the following equations for A, if :

${\sqrt3}$ cot 2 A = 1

Answer

${\sqrt3}$ cot 2 A = 1

⇒ cot 2A = $\dfrac{1}{\sqrt3}$

⇒ cot 2A = cot 60°

So, 2A = 60°

⇒ A = $\dfrac{60°}{2}$

⇒ A = 30°

Hence, A = 30°.

Calculate the value of A, if :

(sin A - 1) (2 cos A - 1) = 0

Answer

(sin A - 1) (2 cos A - 1) = 0

⇒ (sin A - 1) = 0 and (2 cos A - 1) = 0

⇒ sin A = 1 and cos A = $\dfrac{1}{2}$

⇒ sin A = sin 90° and cos A = cos 60°

Hence, A = 90° and 60°.

Calculate the value of A, if :

(tan A - 1) (cosec 3A - 1) = 0

Answer

(tan A - 1) (cosec 3A - 1) = 0

⇒ (tan A - 1) = 0 and (cosec 3A - 1) = 0

⇒ tan A = 1 and cosec 3A = 1

⇒ tan A = tan 45° and cosec 3A = cosec 90°

So, A = 45° and 3A = 90°

⇒ A = 45° and A = 30°

Hence, A = 45° and 30°.

Calculate the value of A, if :

(sec 2A - 1) (cosec 3A - 1) = 0

Answer

(sec 2A - 1) (cosec 3A - 1) = 0

⇒ sec 2A - 1 = 0 and cosec 3A - 1 = 0

⇒ sec 2A = 1 and cosec 3A = 1

⇒ sec 2A = sec 0° and cosec 3A = cosec 90°

So, 2A = 0° and 3A = 90°

⇒ A = 0° and A = $\dfrac{90°}{3} = 30°$

Hence, A = 0° and 30°.

Calculate the value of A, if :

cos 3A (2 sin 2A - 1) = 0

Answer

cos 3A. (2 sin 2A - 1) = 0

⇒ cos 3A = 0 and 2sin 2A - 1 = 0

⇒ cos 3A = 0 and 2sin 2A = 1

⇒ cos 3A = 0 and sin 2A = $\dfrac{1}{2}$

⇒ cos 3A = cos 90° and sin 2A = sin 30°

So, 3A = 90° and 2A = 30°

⇒ A = $\dfrac{90°}{3} = 30°$ and A = $\dfrac{30°}{2} = 15°$

Hence, A = 30° and 15°.

Calculate the value of A, if :

(cosec 2A - 2) (cot 3A - 1) = 0

Answer

(cosec 2A - 2) (cot 3A - 1) = 0

⇒ cosec 2A - 2 = 0 and cot 3A - 1 = 0

⇒ cosec 2A = 2 and cot 3A = 1

⇒ cosec 2A = cosec 30° and cot 3A = cot 45°

So, 2A = 30° and 3A = 45°

⇒ A = $\dfrac{30°}{2} = 15°$ and A = $\dfrac{45°}{3} = 15°$

Hence, A = 15°.

If 2 sin x° - 1 = 0 and x° is an acute angle; find:

(i) sin x°

(ii) x°

(iii) cos x° and tan x°.

Answer

(i) 2 sin x° - 1 = 0

⇒ 2 sin x° = 1

⇒ sin x° = $\dfrac{1}{2}$

Hence, sin x° = $\dfrac{1}{2}$.

(ii) x°

⇒ sin x° = $\dfrac{1}{2}$

⇒ sin x° = sin 30°

Hence, x° = 30°.

(iii) cos x°

⇒ cos 30° = $\dfrac{\sqrt3}{2}$

Hence, cos x° = $\dfrac{\sqrt3}{2}$.

tan x°

⇒ tan 30° = $\dfrac{1}{\sqrt3}$

Hence, tan x° = $\dfrac{1}{\sqrt3}$.

If 4 cos² x° - 1 = 0 and 0 ≤ x° ≤ 90°, find:

(i) x°

(ii) sin² x° + cos² x°

(iii) $\dfrac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°}$

Answer

(i) 4 cos² x° - 1 = 0

⇒ 4 cos² x° = 1

⇒ cos² x° = $\dfrac{1}{4}$

⇒ cos x° = $\sqrt{\dfrac{1}{4}}$

⇒ cos x° = $\dfrac{1}{2}$

⇒ cos x° = cos 60°

Hence, x° = 60°.

(ii) sin² x° + cos² x°

⇒ sin² 60° + cos² 60°

$= \Big(\dfrac{\sqrt3}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{3 + 1}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

Hence, sin² x° + cos² x° = 1.

(iii) $\dfrac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°}$

$= \dfrac{1}{\text{cos}^2 \text{ 60°}} - \text{tan}^2 \text{ 60°}\\[1em] = \dfrac{1}{\Big(\dfrac{1}{2}\Big)^2} - (\sqrt3)^2\\[1em] = \dfrac{4}{1} - 3\\[1em] = 4 - 3\\[1em] = 1$

Hence, $\dfrac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°} = 1$.

If 4 sin² θ - 1 = 0 and angle θ is less than 90°, find the value of θ and hence the value of cos² θ + tan² θ.

Answer

4 sin² θ - 1 = 0

⇒ 4 sin² θ = 1

⇒ sin² θ = $\dfrac{1}{4}$

⇒ sin θ = $\sqrt{\dfrac{1}{4}}$

⇒ sin θ = $\dfrac{1}{2}$

⇒ sin θ = sin 30°

So, θ = 30°

Now, cos² θ + tan² θ

= cos² 30° + tan² 30°

$= \Big(\dfrac{\sqrt3}{2}\Big)^2 + \Big(\dfrac{1}{\sqrt3}\Big)^2\\[1em] = \dfrac{3}{4} + \dfrac{1}{3}\\[1em] = \dfrac{3 \times 3}{4 \times 3} + \dfrac{1 \times 4}{3 \times 4}\\[1em] = \dfrac{9}{12} + \dfrac{4}{12}\\[1em] = \dfrac{9 + 4}{12}\\[1em] = \dfrac{13}{12}\\[1em] = 1\dfrac{1}{12}$

Hence, θ = 30° and cos² 30° + tan² 30° = $\dfrac{13}{12}$ = $1\dfrac{1}{12}$.

If sin 3A = 1 and 0 ≤ A ≤ 90°, find :

(i) sin A

(ii) cos 2 A

(iii) tan² A - $\dfrac{1}{\text{cos}^2 \text{ A}}$

Answer

sin 3A = 1

⇒ sin 3A = sin 90°

So, 3A = 90°

⇒ A = $\dfrac{90°}{3} = 30°$

(i) sin A = sin 30° = $\dfrac{1}{2}$

Hence, sin A = $\dfrac{1}{2}$.

(ii) cos 2 A

= cos (2 x 30°)

= cos 60°

= $\dfrac{1}{2}$

Hence, cos 2A = $\dfrac{1}{2}$.

(iii) tan² A - $\dfrac{1}{\text{cos}^2 \text{ A}}$

$= \text{tan}^2 \text{30°} - \dfrac{1}{\text{cos}^2 \text{30°}}\\[1em] = \Big(\dfrac{1}{\sqrt3}\Big)^2 - \dfrac{1}{\Big(\dfrac{\sqrt3}{2}\Big)^2}\\[1em] = \dfrac{1}{3} - \dfrac{1}{\dfrac{3}{4}}\\[1em] = \dfrac{1}{3} - \dfrac{4}{3}\\[1em] = \dfrac{1 - 4}{3}\\[1em] = \dfrac{- 3}{3}\\[1em] = - 1$

Hence, tan² A - $\dfrac{1}{\text{cos}^2 \text{ A}}$ = -1.

If 2 cos 2A = ${\sqrt3}$ and A is acute, find :

(i) A

(ii) sin 3A

(iii) sin² (75° - A) + cos² (45° + A)

Answer

(i) 2 cos 2A = ${\sqrt3}$

⇒ cos 2A = $\dfrac{\sqrt3}{2}$

⇒ cos 2A = cos 30°

So, 2A = 30°

⇒ A = $\dfrac{30°}{2} = 15°$

Hence, A = 15°.

(ii) sin 3A

= sin (3 x 15°)

= sin 45°

= $\dfrac{1}{\sqrt2}$

Hence, sin 3A = $\dfrac{1}{\sqrt2}$.

(iii) sin² (75° - A) + cos² (45° + A)

= sin² (75° - 15°) + cos² (45° + 15°)

= sin² 60° + cos² 60°

$= \Big(\dfrac{\sqrt3}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{3}{4} + \dfrac{1}{4}\\[1em] = \dfrac{3 + 1}{4} \\[1em] = \dfrac{4}{4} \\[1em] = 1$

Hence, sin² (75° - A) + cos² (45° + A) = 1.