Trigonometrical Ratios — Test Yourself

The value of tan A is :

1. $\dfrac{5}{12}$

2. $\dfrac{12}{13}$

3. $\dfrac{5}{13}$

4. $\dfrac{13}{12}$

Answer

Since ΔPQR is a right angled triangle, using pythagoras theorem,

⇒ Hypotenuse² = Base² + Height²

⇒ PQ² = QR² + PR²

⇒ 13² = PR² + 12²

⇒ 169 = PR² + 144

⇒ PR² = 169 - 144

⇒ PR² = 25

⇒ PR = $\sqrt{25}$

⇒ PR = 5 cm.

By formula, tan A = $\dfrac{\text{Height}}{\text{Base}}$

From figure,

$\text{tan A} = \dfrac{PR}{RQ} \\[1em] = \dfrac{5}{12}.$

Hence, option 1 is the correct option.

The value of sin B - cos A is;

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

2. 1

3. 0

4. none of these

Answer

Since ΔABC is a right angled triangle.

By formula,

sin θ = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$ and cos θ = $\dfrac{\text{Base}}{\text{Hypotenuse}}$.

So, sin B = $\dfrac{\text{AC}}{\text{AB}} = \dfrac{3}{5}$

cos A = $\dfrac{\text{AC}}{\text{AB}} = \dfrac{3}{5}$

The value of sin B - cos A = $\dfrac{3}{5} - \dfrac{3}{5}$ = 0.

Hence, option 3 is the correct option.

The value of cos 60° - sin 90° + 2 cos 0° is :

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

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

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

4. $1\dfrac{1}{2}$

Answer

Solving,

$\Rightarrow \text{cos 60° - sin 90° + 2 cos 0° } = \dfrac{1}{2} - 1 + 2 \times 1\\[1em] = \dfrac{1}{2} - 1 + 2\\[1em] = \dfrac{1}{2} + 1\\[1em] = \dfrac{1 + 2}{2}\\[1em] = \dfrac{3}{2}\\[1em] = 1\dfrac{1}{2}.$

Hence, option 4 is the correct option.

The value of sin 23° - cos 67° is :

1. 1

2. 0

3. cos 44°

4. -cos 44°

Answer

Given,

⇒ sin 23° - cos 67°

⇒ sin 23° - cos (90° - 23°)

⇒ sin 23° - sin 23°

⇒ 0.

Hence, option 2 is the correct option.

Statement 1: The angle C of a right angled triangle is 90°, then tan A = cot B.

Statement 2: Since, angle C of triangle ABC = 90°.

∴ ∠A + ∠B = 90° ⇒ ∠A = 90° - ∠B

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, ∠C = 90°.

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180°

⇒ ∠A + ∠B = 180° - 90°

⇒ ∠A + ∠B = 90°

⇒ ∠A = 90° - ∠B

So, statement 2 is true.

⇒ tan A = tan (90° - ∠B)

⇒ tan A = cot B

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Statement 1: If 4 cos A = 3, sec A = $\dfrac{4}{3}$.

Statement 2: 4 cos A = 3 ⇒ cos A = $\dfrac{3}{4}$ and sec A = $\dfrac{1}{\text{cos A}} = \dfrac{4}{3}$.

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,

⇒ 4 cos A = 3

⇒ cos A = $\dfrac{3}{4}$

⇒ $\dfrac{1}{\text{cos A}} = \dfrac{1}{\dfrac{3}{4}}$

⇒ sec A = $\dfrac{4}{3}$.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Assertion (A): The value of sin² 30° - 2 cos³ 60° + 2 tan⁴ 45° is 2.

Reason (R): sin 30° = $\dfrac{1}{2}$, cos 60° = $\dfrac{1}{2}$, and tan 45° = 1

1. A is true, but R is false.

2. A is false, but R is true.

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

4. Both A and R are true, and R is the incorrect reason for A.

Answer

We know that,

sin 30° = $\dfrac{1}{2}$, cos 60° = $\dfrac{1}{2}$, and tan 45° = 1.

So, reason (R) is true.

$\text{sin}^2 30° - \text{2 cos}^3 60° + \text{2 tan}^4 45° = \Big(\dfrac{1}{2}\Big)^2 - 2 \times \Big(\dfrac{1}{2}\Big)^3 + 2 \times 1^4\\[1em] = \dfrac{1}{4} - 2 \times \dfrac{1}{8} + 2\\[1em] = \dfrac{1}{4} - \dfrac{1}{4} + 2\\[1em] = 2.$

So, assertion (A) is true.

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

Hence, option 3 is the correct option.

Assertion (A): If A = 30°, the value of 4 sin A sin (60° - A) sin (60° + A) = 1.

Reason (R): 60° - A = 30° and 60° + A = 90°.

1. A is true, but R is false.

2. A is false, but R is true.

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

4. Both A and R are true, and R is the incorrect reason for A.

Answer

If A = 30°,

60° - A = 60° - 30° = 30° and 60° + A = 60° + 30° = 90°.

So, reason (R) is true.

sin A = sin 30° = $\dfrac{1}{2}$

sin (60° - A) = sin 30° = $\dfrac{1}{2}$

sin (60° + A) = sin 90° = 1

Substituting values in 4 sin A sin (60° - A) sin (60° + A), we get :

$\Rightarrow 4 \times \dfrac{1}{2} \times \dfrac{1}{2} \times 1\\[1em] \Rightarrow 4 \times \dfrac{1}{4}\\[1em] \Rightarrow 1.$

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

Hence, option 3 is the correct option.

If $\text{cosec θ} = {\sqrt5}$ find the value of :

(i) 2 - sin² θ - cos² θ

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

Answer

Given:

$\text{cosec θ} = {\sqrt5}$

$⇒ \text{cosec θ} = \dfrac{Hypotenuse}{Perpendicular} = {\sqrt5}\\[1em]$

∴ If length of AC = $\sqrt{5}$x unit, length of BC = x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ ($\sqrt{5}$x)² = (x)² + AB²

⇒ 5x² = x² + AB²

⇒ AB² = 5x² - x²

⇒ AB² = 4x²

⇒ AB = $\sqrt{4 \text{x}^2}$

⇒ AB = 2x

(i) sin θ = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BC}{CA} = \dfrac{x}{\sqrt{5}x} = \dfrac{1}{\sqrt{5}}$

cos θ = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC} = \dfrac{2x}{\sqrt{5}x} = \dfrac{2}{\sqrt{5}}$

Now,

2 - sin²θ - cos²θ

$= 2 - \Big(\dfrac{1}{\sqrt{5}}\Big)^2 - \Big(\dfrac{2}{\sqrt{5}}\Big)^2\\[1em] = 2 - \dfrac{1}{5} - \dfrac{4}{5}\\[1em] = 2 + \dfrac{-1 - 4}{5}\\[1em] = 2 + \dfrac{-5}{5}\\[1em] = 2 - 1\\[1em] = 1$

Hence, 2 - sin²θ - cos²θ = 1.

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

$= 2 + \dfrac{1}{\Big(\dfrac{1}{\sqrt{5}}\Big)^2} - \dfrac{\Big(\dfrac{2}{\sqrt{5}}\Big)^2}{\Big(\dfrac{1}{\sqrt{5}}\Big)^2}\\[1em] = 2 + \dfrac{1}{\dfrac{1}{5}} - \dfrac{\dfrac{4}{5}}{\dfrac{1}{5}}\\[1em] = 2 + \dfrac{5}{1} - \dfrac{\dfrac{4}{\cancel{5}}}{\dfrac{1}{\cancel{5}}}\\[1em] = 2 + 5 - \dfrac{4}{1}\\[1em] = 7 - 4\\[1em] = 3$

Hence, $2 + \dfrac{1}{\text{sin}^2 \text{ θ}} - \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}$ = 3.

In the given figure; ∠C = 90° and D is mid-point of AC. Find :

(i) $\dfrac{\text{tan ∠CAB}}{\text{tan ∠CDB}}$

(ii) $\dfrac{\text {tan ∠ABC}}{\text {tan ∠DBC}}$

Answer

Since D is the mid-point of A. So, AC = 2DC

(i) $\text{tan ∠CAB} = \dfrac{Perpendicular}{Base} = \dfrac{BC}{AC}$

$\text{tan ∠CDB} = \dfrac{Perpendicular}{Base} = \dfrac{BC}{DC}$

Now,

$\dfrac{\text{tan ∠CAB}}{\text{tan ∠CDB}}\\[1em] = \dfrac{\dfrac{BC}{AC}}{\dfrac{BC}{DC}}\\[1em] = \dfrac{BC \times DC}{AC \times BC}\\[1em] = \dfrac{\cancel{BC} \times DC}{AC \times \cancel{BC}}\\[1em] = \dfrac{DC}{AC}\\[1em] = \dfrac{DC}{2 \times DC}\\[1em] = \dfrac{\cancel{DC}}{2 \times \cancel{DC}}\\[1em] = \dfrac{1}{2}$

Hence, $\dfrac{\text{tan ∠CAB}}{\text{tan ∠CDB}} = \dfrac{1}{2}$

(ii) $\text{tan ∠ABC} = \dfrac{Perpendicular}{Base} = \dfrac{AC}{BC}$

$\text{tan ∠DBC} = \dfrac{Perpendicular}{Base} = \dfrac{DC}{BC}$

Now,

$\dfrac{\text {tan ∠ABC}}{\text {tan ∠DBC}}\\[1em] = \dfrac{\dfrac{AC}{BC}}{\dfrac{DC}{BC}}\\[1em] = \dfrac{AC \times BC}{BC \times DC}\\[1em] = \dfrac{AC \times \cancel{BC}}{\cancel{BC} \times DC}\\[1em] = \dfrac{AC}{DC}\\[1em] = \dfrac{2 \times DC}{DC}\\[1em] = \dfrac{2 \times \cancel{DC}}{\cancel{DC}}\\[1em] = 2$

Hence, $\dfrac{\text {tan ∠ABC}}{\text {tan ∠DBC}} = 2$

If 3 cos A = 4 sin A, find the value of :

(i) cos A

(ii) 3 - cot² A + cosec² A

Answer

Given:

3 cos A = 4 sin A

$\dfrac{\text{cos A}}{\text{sin A}} = \dfrac{4}{3}$

$\text{cot A} = \dfrac{4}{3}$

$\text{cot A} = \dfrac{Base}{Perpendicular} = \dfrac{4}{3}$

∴ If length of AB = 4x unit, length of BC = 3x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (4x)² + (3x)²

⇒ AC² = 16x² + 9x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25\text{x}^2}$

⇒ AC = 5x

(i) cos $A = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$

Hence, cos $A = \dfrac{4}{5}$.

(ii) 3 - cot² A + cosec² A

cot $A = \dfrac{Base}{Perpendicular}$

$= \dfrac{AB}{BC} = \dfrac{4x}{3x} = \dfrac{4}{3}$

cosec $A = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AC}{BC} = \dfrac{5x}{3x} = \dfrac{5}{3}$

Now,

3 - cot² A + cosec² A

$= 3 - \Big(\dfrac{4}{3}\Big)^2 + \Big(\dfrac{5}{3}\Big)^2\\[1em] = 3 - \dfrac{16}{9} + \dfrac{25}{9}\\[1em] = 3 + \dfrac{-16 + 25}{9}\\[1em] = 3 + \dfrac{9}{9}\\[1em] = 3 + 1\\[1em] = 4$

Hence, 3 - cot² A + cosec² A = 4.

Use the information given in the following figure to evaluate :

$\dfrac{10}{\text{sin x}} + \dfrac{6}{\text{sin y}} - \text{6 cot y}$.

Answer

From the figure, in Δ ADC,

⇒ AC² = DC² + AD² (∵ AC is hypotenuse)

⇒ 20² = DC² + 12²

⇒ 400 = DC² + 144

⇒ DC² = 400 - 144

⇒ DC² = 256

⇒ DC = $\sqrt{256}$

⇒ DC = 16

BD = BC - DC

= 21 - 16 = 5

In Δ ABD,

⇒ AB² = AD² + BD² (∵ AB is hypotenuse)

⇒ AB² = 12² + 5²

⇒ AB² = 144 + 25

⇒ AB² = 169

⇒ AB = $\sqrt{169}$

⇒ AB = 13

sin x = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BD}{AB} = \dfrac{5}{13}$

sin y = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{AD}{AC} = \dfrac{12}{20} = \dfrac{3}{5}$

cot y = $\dfrac{Base}{Perpendicular}$

$= \dfrac{DC}{AD} = \dfrac{16}{12} = \dfrac{4}{3}$

Now,

$\dfrac{10}{\text{sin x}} + \dfrac{6}{\text{sin y}} - \text{6 cot y}\\[1em] = \dfrac{10}{\dfrac{5}{13}} + \dfrac{6}{\dfrac{3}{5}} - 6 \times \dfrac{4}{3}\\[1em] = \dfrac{10 \times 13}{5} + \dfrac{6 \times 5}{3} - \dfrac{24}{3}\\[1em] = \dfrac{130}{5} + \dfrac{30}{3} - \dfrac{24}{3}\\[1em] = 26 + 10 - 8\\[1em] = 28$

Hence, $\dfrac{10}{\text{sin x}} + \dfrac{6}{\text{sin y}} - \text{6 cot y} = 28.$

Evaluate :

$\dfrac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}}$, when A = 15°.

Answer

$\dfrac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}}\\[1em] = \dfrac{\text{cos (3 x 15°) - 2 cos (4 x 15°)}}{\text{sin (3 x 15°) + 2 sin (4 x 15°)}}\\[1em] = \dfrac{\text{cos 45° - 2 cos 60°}}{\text{sin 45° + 2 sin 60°}}\\[1em] = \dfrac{\Big(\dfrac{1}{\sqrt2}\Big) - 2 \times \Big(\dfrac{1}{2}\Big)}{\Big(\dfrac{1}{\sqrt2}\Big) + 2 \times \Big(\dfrac{\sqrt3}{2}\Big)}\\[1em] = \dfrac{\dfrac{1}{\sqrt2} - 1}{\dfrac{1}{\sqrt2} + \sqrt3}\\[1em] = \dfrac{\dfrac{1}{\sqrt2} - \dfrac{1 \times \sqrt2}{\sqrt2}}{\dfrac{1}{\sqrt2} + \dfrac{\sqrt3 \times \sqrt2}{\sqrt2}}\\[1em] = \dfrac{\dfrac{1}{\sqrt2} - \dfrac{\sqrt2}{\sqrt2}}{\dfrac{1}{\sqrt2} + \dfrac{\sqrt6}{\sqrt2}}\\[1em] = \dfrac{\dfrac{1 - \sqrt2}{\sqrt2}}{\dfrac{1 + \sqrt6}{\sqrt2}}\\[1em] = \dfrac{\dfrac{1 - \sqrt2}{\cancel{\sqrt2}}}{\dfrac{1 + \sqrt6}{\cancel{\sqrt2}}}\\[1em] = \dfrac{1 - \sqrt2}{1 + \sqrt6}\\[1em] = \dfrac{(1 - \sqrt2) \times (1 - \sqrt6)}{(1 + \sqrt6) \times (1 - \sqrt6)}\\[1em] = \dfrac{1 - \sqrt2 - \sqrt6 + \sqrt{12}}{1 - 6}\\[1em] = \dfrac{1 - \sqrt2 - \sqrt6 + 2\sqrt3}{-5}\\[1em] = \dfrac{1}{5}(- 1 + \sqrt2 + \sqrt6 - 2\sqrt3)$

Hence, $\dfrac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}} = \dfrac{1}{5}(- 1 + \sqrt2 + \sqrt6 - 2\sqrt3)$

Evaluate :

$\dfrac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}}$; when B = 20°.

Answer

$\dfrac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}}\\[1em] = \dfrac{\text{3 sin (3 x 20°) + 2 cos((2 x 20°) + 5°)}}{\text{2 cos (3 x 20°) - sin((2 x 20°) - 10°)}}\\[1em] = \dfrac{\text{3 sin 60° + 2 cos(40° + 5°)}}{\text{2 cos 60° - sin(40° - 10°)}}\\[1em] = \dfrac{\text{3 sin 60° + 2 cos 45°}}{\text{2 cos 60° - sin 30°}}\\[1em] = \dfrac{3 \times \dfrac{\sqrt3}{2} + 2 \times \dfrac{1}{\sqrt2}}{2 \times \dfrac{1}{2} - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3}{2} + \dfrac{2}{\sqrt2}}{\dfrac{2}{2} - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3}{2} + \dfrac{2 \times \sqrt2}{\sqrt2 \times \sqrt2}}{\dfrac{2 - 1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3}{2} + \dfrac{2\sqrt2}{2}}{\dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3 + 2\sqrt2}{2}}{\dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3 + 2\sqrt2}{\cancel2}}{\dfrac{1}{\cancel2}}\\[1em] = 3\sqrt3 + 2\sqrt2$

Hence, $\dfrac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}} = 3\sqrt3 + 2\sqrt2.$

Solve for x :

2 cos 3x - 1 = 0

Answer

2 cos 3x - 1 = 0

⇒ 2 cos 3x = 1

⇒ cos 3x = $\dfrac{1}{2}$

⇒ cos 3x = cos 60°

So, 3x = 60°

⇒ x = $\dfrac{60°}{3} = 20°$

Hence, x = 20°.

Solve for x :

cos $\dfrac{x}{3} - 1 = 0$

Answer

$\text{cos }\dfrac{x}{3} - 1 = 0$

$⇒ \text{cos } \dfrac{x}{3} = 1\\[1em] ⇒ \text{cos } \dfrac{x}{3} = \text{cos 0°}$

So,

$⇒ \dfrac{x}{3} = 0°\\[1em] ⇒ x = 3 \times 0°\\[1em] ⇒ x = 0°\\[1em]$

Hence, x = 0°.

Solve for x :

sin (x + 10°) = $\dfrac{1}{2}$

Answer

sin (x + 10°) = $\dfrac{1}{2}$

⇒ sin (x + 10°) = sin 30°

So, x + 10° = 30°

⇒ x = 30° - 10° = 20°

Hence, x = 20°.

Solve for x :

cos (2x - 30°) = 0

Answer

cos (2x - 30°) = 0

⇒ cos (2x - 30°) = cos 90°

So, 2x - 30° = 90°

⇒ 2x = 90° + 30°

⇒ x = $\dfrac{120°}{2}$

⇒ x = 60°

Hence, x = 60°.

Solve for x :

2 cos (3x - 15°) = 1

Answer

2 cos (3x - 15°) = 1

⇒ cos (3x - 15°) = $\dfrac{1}{2}$

⇒ cos (3x - 15°) = cos 60°

So, 3x - 15° = 60°

⇒ 3x = 60° + 15°

⇒ x = $\dfrac{75°}{3}$

⇒ x = 25°

Hence, x = 25°.

Solve for x :

tan² (x - 5°) = 3

Answer

tan² (x - 5°) = 3

⇒ tan (x - 5°) = $\sqrt3$

⇒ tan (x - 5°) = tan 60°

So, x - 5° = 60°

⇒ x = 60° + 5°

⇒ x = 65°

Hence, x = 65°.

Solve for x :

3 tan² (2x - 20°) = 1

Answer

3 tan² (2x - 20°) = 1

⇒ tan² (2x - 20°) = $\dfrac{1}{3}$

⇒ tan (2x - 20°) = $\sqrt\dfrac{1}{3}$

⇒ tan (2x - 20°) = $\dfrac{1}{\sqrt3}$

⇒ tan (2x - 20°) = tan 30°

So, 2x - 20° = 30°

⇒ 2x = 30° + 20°

⇒ 2x = 50°

⇒ x = $\dfrac{50°}{2}$

⇒ x = 25°

Hence, x = 25°.

Solve for x :

$cos\Big(\dfrac{x}{2} + 10°\Big)$ = $\dfrac{\sqrt3}{2}$

Answer

$\text{cos}\Big(\dfrac{x}{2} + 10°\Big)$ = $\dfrac{\sqrt3}{2}$

$⇒ \text{cos } \Big(\dfrac{x}{2} + 10°\Big) = \text{cos 30°}$

So,

$⇒ \Big(\dfrac{x}{2} + 10°\Big) = 30°\\[1em] ⇒ \dfrac{x}{2} = 30° - 10°\\[1em] ⇒ x = 20° \times 2\\[1em] ⇒ x = 40° \\[1em]$

Hence, x = 40°.

Solve for x :

sin² x + sin² 30° = 1

Answer

sin² x + sin² 30° = 1

$⇒ \text{sin }^2 x + \Big(\dfrac{1}{2}\Big)^2 = 1\\[1em] ⇒ \text{sin }^2 x + \Big(\dfrac{1}{4}\Big) = 1\\[1em] ⇒ \text{sin }^2 x = 1 - \Big(\dfrac{1}{4}\Big)\\[1em] ⇒ \text{sin }^2 x = \dfrac{4}{4} - \dfrac{1}{4}\\[1em] ⇒ \text{sin }^2 x = \dfrac{4 - 1}{4}\\[1em] ⇒ \text{sin }^2 x = \dfrac{3}{4}\\[1em] ⇒ \text{sin } x = \sqrt\dfrac{3}{4}\\[1em] ⇒ \text{sin } x = \dfrac{\sqrt3}{2}\\[1em] ⇒ \text{sin } x = \text{sin } 60°\\[1em]$

So, x = 60°

Hence, x = 60°.

Solve for x :

cos² 30° + cos² x = 1

Answer

cos² 30° + cos² x = 1

$⇒ \Big(\dfrac{\sqrt{3}}{2}\Big)^2 + \text{cos }^2 x = 1\\[1em] ⇒ \Big(\dfrac{3}{4}\Big) + \text{cos }^2 x = 1\\[1em] ⇒ \text{cos }^2 x = 1 - \Big(\dfrac{3}{4}\Big)\\[1em] ⇒ \text{cos }^2 x = \dfrac{4}{4} - \dfrac{3}{4}\\[1em] ⇒ \text{cos }^2 x = \dfrac{4 - 3}{4}\\[1em] ⇒ \text{cos }^2 x = \dfrac{1}{4}\\[1em] ⇒ \text{cos } x = \sqrt\dfrac{1}{4}\\[1em] ⇒ \text{cos } x = \dfrac{1}{2}\\[1em] ⇒ \text{cos } x = \text{cos 60°} \\[1em]$

So, x = 60°

Hence, x = 60°.

Solve for x :

cos² 30° + sin² 2x = 1

Answer

cos² 30° + sin² 2x = 1

$⇒ \Big(\dfrac{\sqrt3}{2}\Big)^2 + \text{sin }^2 2x = 1\\[1em] ⇒ \Big(\dfrac{3}{4}\Big) + \text{sin }^2 2x = 1\\[1em] ⇒ \text{sin}^2 2x = 1 - \dfrac{3}{4}\\[1em] ⇒ \text{sin}^2 2x = \dfrac{4}{4} - \dfrac{3}{4}\\[1em] ⇒ \text{sin}^2 2x = \dfrac{4 - 3}{4}\\[1em] ⇒ \text{sin}^2 2x = \dfrac{1}{4}\\[1em] ⇒ \text{sin} 2x = \sqrt\dfrac{1}{4}\\[1em] ⇒ \text{sin} 2x = \dfrac{1}{2}\\[1em] ⇒ \text{sin} 2x = \text{sin }30°$

So, 2x = 30°

⇒ x = $\dfrac{30°}{2}$

⇒ x = 15°

Hence, x = 15°.

Solve for x :

sin² 60° + cos² (3x - 9°) = 1

Answer

sin² 60° + cos² (3x - 9°) = 1

$⇒ \Big(\dfrac{\sqrt3}{2}\Big)^2 + \text{cos }^2 (3x - 9°) = 1\\[1em] ⇒ \Big(\dfrac{3}{4}\Big) + \text{cos }^2 (3x - 9°) = 1\\[1em] ⇒ \text{cos}^2 (3x - 9°) = 1 - \dfrac{3}{4}\\[1em] ⇒ \text{cos}^2 (3x - 9°) = \dfrac{4}{4} - \dfrac{3}{4}\\[1em] ⇒ \text{cos}^2 (3x - 9°) = \dfrac{4 - 3}{4} \\[1em] ⇒ \text{cos}^2 (3x - 9°) = \dfrac{1}{4} \\[1em] ⇒ \text{cos} (3x - 9°) = \sqrt\dfrac{1}{4} \\[1em] ⇒ \text{cos} (3x - 9°) = \dfrac{1}{2} \\[1em] ⇒ \text{cos} (3x - 9°) = \text{cos } 60°$

So, 3x - 9° = 60°

⇒ 3x = 60° + 9°

⇒ 3x = 69°

⇒ x = $\dfrac{69°}{3}$

⇒ x = 23°

Hence, x = 23°.

If 2 cos (A + B) = 2 sin (A - B) = 1; find the values of A and B.

Answer

2 cos (A + B) = 1

⇒ cos (A + B) = $\dfrac{1}{2}$

⇒ cos (A + B) = cos 60°

So, A + B = 60° ...............(1)

2 sin (A - B) = 1

⇒ sin (A - B) = $\dfrac{1}{2}$

⇒ sin (A - B) = sin 30°

So, A - B = 30° ...............(2)

Adding equation (1) and (2), we get

(A + B) + (A - B) = 60° + 30°

⇒ A + B + A - B = 90°

⇒ 2A = 90°

⇒ A = $\dfrac{90°}{2}$

⇒ A = 45°

From equation (2), A - B = 30°

⇒ 45° - B = 30°

⇒ B = 45° - 30°

⇒ B = 15°

Hence, A = 45° and B = 15°.

For the triangle ABC, show that

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

Answer

For triangle ABC,

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ B + ∠ C = 180° - ∠ A

⇒ $\dfrac{B + C}{2} = \dfrac{180° - A}{2}$

⇒ $\dfrac{B + C}{2} = 90° - \dfrac{A}{2}$

$\text{L.H.S.} = \text{sin}^2\dfrac{A}{2} + \text{sin}^2\dfrac{B+C}{2}\\[1em] = \text{sin}^2\dfrac{A}{2} + \text{sin}^2\Big(90° - \dfrac{A}{2}\Big)\\[1em] = \text{sin}^2\dfrac{A}{2} + \text{cos}^2\dfrac{A}{2}\\[1em] = 1$

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence, $\text{sin}^2\dfrac{A}{2}$ + $\text{sin}^2\dfrac{B+C}{2}$ = 1.

If sec (90° - 3A).cos 48° = 1 and 0 ≤ 3A ≤ 90°; find the value of angle A.

Answer

Given:

sec (90° - 3A) . cos 48° = 1

⇒ cosec 3A . cos 48° = 1

⇒ $\dfrac{1}{\text{sin 3A}}$ . cos 48° = 1

⇒ sin 3A = cos 48°

⇒ sin 3A = cos (90° - 42°)

⇒ sin 3A = sin 42°

So, 3A = 42°

⇒ A = $\dfrac{42°}{3}$

⇒ A = 14°

Hence, A = 14°.

In △ABC, angle C is 90° then find the value of sin (A + B).

Answer

In △ ABC,

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ A + ∠ B + 90° = 180°

⇒ ∠ A + ∠ B = 180° - 90°

⇒ ∠ A + ∠ B = 90°

⇒ sin(A + B) = sin 90°

⇒ sin(A + B) = 1

Hence, sin (A + B) = 1.

If sec A sin A = 0, find the value of cos A.

Answer

sec A sin A = 0

⇒ $\dfrac{1}{\text{cos A}}$ sin A = 0

⇒ tan A = 0

⇒ tan A = tan 0°

Thus, A = 0°.

Now, cos A = cos 0° = 1

Hence, cos A = 1.

Find angle A, if sec 2A = cosec (A + 48°).

Answer

We know that,

sec θ = cosec (90° - θ)

Solving,

⇒ sec 2A = cosec (A + 48°)

⇒ cosec (90° - 2A) = cosec (A + 48°)

⇒ 90° - 2A = A + 48°

⇒ A + 2A = 90° - 48°

⇒ 3A = 42°

⇒ A = $\dfrac{42°}{3}$ = 14°.

Hence, A = 14°.