Trigonometrical Ratios — Exercise 21(C)

$\dfrac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$ is equal to :

1. sin 60°

2. cos 60°

3. sec 60°

4. cosec 60°

Answer

$\dfrac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}} = \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 + \Big(\dfrac{1}{\sqrt{3}}\Big)^2}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{2 \times 3}{4 \times {\sqrt{3}}}\\[1em] = \dfrac{6}{4{\sqrt{3}}}\\[1em] = \dfrac{3}{2{\sqrt{3}}}\\[1em] = \dfrac{{\sqrt{3}}}{2}\\[1em] = \text{sin} 60°$

Hence, option 1 is the correct option.

If tan 3A - ${\sqrt3}$ = 0 and 0 ≤ 3A ≤ 90° ; the measure of angle A is :

1. 15°

2. 20°

3. 30°

4. 10°

Answer

tan 3A - ${\sqrt3}$ = 0

tan 3A = ${\sqrt3}$

tan 3A = tan 60°

So, 3A = 60°

A = $\dfrac{60°}{3}$

A = 20°

Hence, option 2 is the correct option.

If cot A = tan A and 0 ≤ A ≤ 90°, the measure of angle A is :

1. 30°

2. 60°

3. 45°

4. 90°

Answer

cot A = tan A

⇒ $\dfrac{1}{\text{tan A}}$ = tan A

⇒ tan² A = 1

⇒ tan A = 1

⇒ tan A = tan 45°

Hence, A = 45°

Hence, option 3 is the correct option.

The value of :

cos² 60° - 2 sin³ 30° + 3 cot⁴ 45° is :

1. 1

2. -2

3. 3

4. 2

Answer

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

Hence, option 3 is the correct option.

The value of $\dfrac{\text{cos 60° - cos 0°} + \text{2 sin 90°}}{\text{ cot 60°}\times \text{ cot 30°}}$ is :

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

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

3. -2

4. 1

Answer

$\dfrac{\text{cos 60° - cos 0°} + \text{2 sin 90°}}{\text{ cot 60°}\times \text{ cot 30°}} = \dfrac{\dfrac{1}{2} - 1 + 2 \times 1}{\dfrac{1}{\sqrt3}\times \sqrt3}\\[1em] = \dfrac{\dfrac{1}{2} - 1 + 2}{1}\\[1em] = \dfrac{1}{2} + 1\\[1em] = \dfrac{1 + 2}{2}\\[1em] = \dfrac{3}{2}\\[1em] = 1\dfrac{1}{2}$

Hence, option 1 is the correct option.

Find the value of:

sin 30° cos 30°

Answer

$\text{sin 30° cos 30°} = \dfrac{1}{2} \times \dfrac{\sqrt3}{2}\\[1em] = \dfrac{\sqrt3}{4}\\[1em]$

Hence, sin 30° cos 30° = $\dfrac{\sqrt3}{4}$.

Find the value of:

tan 30° tan 60°

Answer

$\text{tan 30° tan 60°} = \dfrac{1}{\sqrt3} \times \sqrt3\\[1em] = \dfrac{1}{\cancel{\sqrt3}} \times \cancel{\sqrt3}\\[1em] = 1$

Hence, tan 30° tan 60° = 1.

Find the value of:

cos² 60° + sin² 30°

Answer

$\text{cos}^2 60° + \text{sin}^2 30° = \Big(\dfrac{1}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{1}{4} + \dfrac{1}{4}\\[1em] = \dfrac{1 + 1}{4}\\[1em] = \dfrac{2}{4}\\[1em] = \dfrac{1}{2}$

Hence, cos² 60° + sin² 30° = $\dfrac{1}{2}$.

Find the value of:

cosec² 60° - tan² 30°

Answer

$\text{cosec}^2 60° - \text{tan}^2 30° = \Big(\dfrac{2}{\sqrt3}\Big)^2 - \Big(\dfrac{1}{\sqrt3}\Big)^2\\[1em] = \dfrac{4}{3} - \dfrac{1}{3}\\[1em] = \dfrac{4 - 1}{3}\\[1em] = \dfrac{3}{3}\\[1em] = 1$

Hence, cosec² 60° - tan² 30° = 1.

Find the value of:

sin² 30° + cos² 30° + cot² 45°

Answer

$\text{sin}^2 30° + \text{cos}^2 30° + \text{cot}^2 45° = \Big(\dfrac{1}{2}\Big)^2 + \Big(\dfrac{\sqrt3}{2}\Big)^2 + (1)^2\\[1em] = \dfrac{1}{4} + \dfrac{3}{4} + 1\\[1em] = \dfrac{1 + 3}{4} + 1\\[1em] = \dfrac{4}{4} + 1\\[1em] = 1 + 1\\[1em] = 2$

Hence, sin² 30° + cos² 30° + cot² 45° = 2.

Find the value of:

cos² 60° + sec² 30° + tan² 45°

Answer

$\text{cos}^2 60° + \text{sec}^2 30° + \text{tan}^2 45° = \Big(\dfrac{1}{2}\Big)^2 + \Big(\dfrac{2}{\sqrt3}\Big)^2 + (1)^2\\[1em] = \dfrac{1}{4} + \dfrac{4}{3} + 1\\[1em] = \dfrac{1 \times 3}{4 \times 3} + \dfrac{4 \times 4}{3 \times 4} + \dfrac{1 \times 12}{12}\\[1em] = \dfrac{3}{12} + \dfrac{16}{12} + \dfrac{12}{12}\\[1em] = \dfrac{3 + 16 + 12}{12}\\[1em] = \dfrac{31}{12}\\[1em] = 2\dfrac{7}{12}$

Hence, cos² 60° + sec² 30° + tan² 45° = $2\dfrac{7}{12}$.

Find the value of :

tan² 30° + tan² 45° + tan² 60°

Answer

tan² 30° + tan² 45° + tan² 60°

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

Hence, tan² 30° + tan² 45° + tan² 60° = $4\dfrac{1}{3}$.

Find the value of :

$\dfrac{\text{tan 45°}}{\text{cosec 30°}} + \dfrac{\text{sec 60°}}{\text{cot 45°}} - \dfrac{\text{5 sin 90°}}{\text{2 cos 0°}}$

Answer

$\dfrac{\text{tan 45°}}{\text{cosec 30°}} + \dfrac{\text{sec 60°}}{\text{cot 45°}} - \dfrac{\text{5 sin 90°}}{\text{2 cos 0°}}$

$= \dfrac{1}{2} + \dfrac{2}{1} - \dfrac{5 \times 1}{2 \times 1}\\[1em] = \dfrac{1}{2} + \dfrac{2 \times 2}{1 \times 2} - \dfrac{5}{2}\\[1em] = \dfrac{1}{2} + \dfrac{4}{2} - \dfrac{5}{2}\\[1em] = \dfrac{1 + 4 - 5}{2}\\[1em] = 0$

Hence, $\dfrac{\text{tan 45°}}{\text{cosec 30°}} + \dfrac{\text{sec 60°}}{\text{cot 45°}} - \dfrac{\text{5 sin 90°}}{\text{2 cos 0°}} = 0$

Find the value of :

3 sin² 30° + 2 tan² 60° - 5 cos² 45°

Answer

3 sin² 30° + 2 tan² 60° - 5 cos² 45°

$= 3 \times \Big(\dfrac{1}{2}\Big)^2 + 2 \times (\sqrt3)^2 - 5 \times \Big(\dfrac{1}{\sqrt2}\Big)^2\\[1em] = 3 \times \dfrac{1}{4} + 2 \times 3 - 5 \times \dfrac{1}{2}\\[1em] = \dfrac{3}{4} + 6 - \dfrac{5}{2}\\[1em] = \dfrac{3}{4} + \dfrac{6 \times 4}{4} - \dfrac{5 \times 2}{2 \times 2}\\[1em] = \dfrac{3}{4} + \dfrac{24}{4} - \dfrac{10}{4}\\[1em] = \dfrac{3 + 24 - 10}{4}\\[1em] = \dfrac{17}{4}\\[1em] = 4\dfrac{1}{4}$

Hence, 3 sin² 30° + 2 tan² 60° - 5 cos² 45° = $4\dfrac{1}{4}$.

Prove that :

sin 60° cos 30° + cos 60°. sin 30° = 1

Answer

sin 60° cos 30° + cos 60°. sin 30° = 1

L.H.S = sin 60° cos 30° + cos 60°. sin 30°

$= \dfrac{\sqrt3}{2} \times \dfrac{\sqrt3}{2} + \dfrac{1}{2} \times \dfrac{1}{2} \\[1em] = \dfrac{3}{4} + \dfrac{1}{4}\\[1em] = \dfrac{3 + 1}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

R.H.S = 1

∴ L.H.S = R.H.S

Hence proved, sin 60° cos 30° + cos 60°. sin 30° = 1.

Prove that :

cos 30°. cos 60° - sin 30°. sin 60° = 0

Answer

cos 30°. cos 60° - sin 30°. sin 60° = 0

L.H.S. = cos 30°. cos 60° - sin 30°. sin 60°

$= \dfrac{\sqrt3}{2} \times \dfrac{1}{2} - \dfrac{1}{2} \times \dfrac{\sqrt3}{2} \\[1em] = \dfrac{\sqrt3}{4} - \dfrac{\sqrt3}{4}\\[1em] = 0$

R.H.S. = 0

∴ L.H.S. = R.H.S.

Hence proved, cos 30°. cos 60° - sin 30°. sin 60° = 0.

Prove that :

cosec² 45° - cot² 45° = 1

Answer

cosec² 45° - cot² 45° = 1

L.H.S. = cosec² 45° - cot² 45°

$= (\sqrt2)^2 - (1)^2 \\[1em] = 2 - 1\\[1em] = 1$

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence proved, cosec² 45° - cot² 45° = 1.

Prove that :

cos² 30° - sin² 30° = cos 60°

Answer

cos² 30° - sin² 30° = cos 60°

L.H.S. = cos² 30° - sin² 30°

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

R.H.S.. = cos 60° = $\dfrac{1}{2}$

∴ L.H.S. = R.H.S.

Hence proved, cos² 30° - sin² 30° = cos 60°.

Prove that :

$\Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

Answer

$\Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

$\text{L.H.S.} = \Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2\\[1em] = \Big(\dfrac{\sqrt3 + 1}{\sqrt3 - 1}\Big)^2\\[1em] = \Big(\dfrac{(\sqrt3 + 1) \times (\sqrt3 + 1)}{(\sqrt3 - 1) \times (\sqrt3 + 1)}\Big)^2\\[1em] = \Big(\dfrac{(\sqrt3 + 1)^2}{(\sqrt3)^2 - (1)^2}\Big)^2\\[1em] = \Big(\dfrac{3 + 1 + 2 \times 1 \times \sqrt3}{3 - 1}\Big)^2\\[1em] = \Big(\dfrac{4 + 2\sqrt3}{2}\Big)^2\\[1em] = (2 + \sqrt3)^2\\[1em] = 4 + 3 + 2 \times 2 \times \sqrt3\\[1em] = 7 + 4\sqrt3$

$\text{R.H.S.} = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}\\[1em] = \dfrac{1 + \dfrac{\sqrt3}{2}}{1 - \dfrac{\sqrt3}{2}}\\[1em] = \dfrac{\dfrac{2 + \sqrt3}{2}}{\dfrac{2 - \sqrt3}{2}}\\[1em] = \dfrac{\dfrac{2 + \sqrt3}{\cancel{2}}}{\dfrac{2 - \sqrt3}{\cancel{2}}}\\[1em] = \dfrac{2 + \sqrt3}{2 - \sqrt3}\\[1em] = \dfrac{(2 + \sqrt3) \times (2 + \sqrt3)}{(2 - \sqrt3) \times (2 + \sqrt3)}\\[1em] = \dfrac{(2 + \sqrt3)^2}{(2)^2 - (\sqrt3)^2}\\[1em] = \dfrac{4 + 3 + 2 \times 2 \times \sqrt3}{4 - 3}\\[1em] = 7 + 4\sqrt3\\[1em]$

∴ L.H.S. = R.H.S.

Hence, $\Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

Prove that :

3 cosec² 60° - 2 cot² 30° + sec² 45° = 0.

Answer

3 cosec² 60° - 2 cot² 30° + sec² 45° = 0.

L.H.S. = 3 cosec² 60° - 2 cot² 30° + sec² 45°

$= 3 \times \Big(\dfrac{2}{\sqrt3}\Big)^2 - 2 \times ({\sqrt3})^2 + ({\sqrt2})^2\\[1em] = 3 \times \Big(\dfrac{4}{3}\Big) - 2 \times 3 + 2\\[1em] = 4 - 6 + 2\\[1em] = 0$

R.H.S. = 0

∴ L.H.S. = R.H.S.

Hence, 3 cosec² 60° - 2 cot² 30° + sec² 45° = 0.

Prove that :

$\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

Answer

$\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

L.H.S. = sin (2 x 30°) = sin 60° = $\dfrac{\sqrt3}{2}$

R.H.S.

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

∴ L.H.S. = R.H.S.

Hence, $\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

Prove that :

$\text{cos }(2 \times 30°) = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}$

Answer

$\text{cos }(2 \times 30°) = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}$

L.H.S. = cos (2 x 30°) = cos 60° = $\dfrac{1}{2}$

R.H.S.

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

∴ L.H.S. = R.H.S.

Hence, $\text{cos }(2 \times 30°) = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}$

Prove that :

$\text{tan} (2 \times 30°) = \dfrac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}$

Answer

$\text{tan} (2 \times 30°) = \dfrac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}$

L.H.S. = tan 2 x 30° = tan 60° = $\sqrt3$

R.H.S.

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

∴ L.H.S. = R.H.S.

Hence, $\text{tan} (2 \times 30°) = \dfrac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}$

ABC is an isosceles right-angled triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios :

(i) sin 45°

(ii) cos 45°

(iii) tan 45°

Answer

In Δ ABC,

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

⇒ AC² = x² + x²

⇒ AC² = 2x²

⇒ AC = $\sqrt{2\text{x}^2}$

⇒ AC = $\sqrt{2}$ x

As Δ ABC is isosceles right angled triangle, ∠BAC = ∠BCA = $\dfrac{90°}{2}$ = 45°

(i) sin 45°

sin 45° = sin A = sin C

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

= $\dfrac{BC}{AC} = \dfrac{x}{\sqrt2x} = \dfrac{1}{\sqrt2}$

Hence, sin 45° = $\dfrac{1}{\sqrt2}$.

(ii) cos 45°

cos 45° = cos A = cos C

cos A = $\dfrac{Base}{Hypotenuse}$

= $\dfrac{AB}{AC} = \dfrac{x}{\sqrt2x} = \dfrac{1}{\sqrt2}$

Hence, cos 45° = $\dfrac{1}{\sqrt2}$.

(iii) tan 45°

tan 45° = tan A = tan C

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

= $\dfrac{BC}{AB} = \dfrac{x}{x}$ = 1

Hence, tan 45° = 1.

Prove that :

sin 60° = 2 sin 30° cos 30°.

Answer

sin 60° = 2 sin 30° cos 30°.

L.H.S. = sin 60° = $\dfrac{\sqrt3}{2}$

R.H.S.= 2 sin 30° cos 30°

$= 2 \times \dfrac{1}{2} \times \dfrac{\sqrt3}{2}\\[1em] = \cancel{2} \times \dfrac{1}{\cancel{2}} \times \dfrac{\sqrt3}{2}\\[1em] = \dfrac{\sqrt3}{2}$

∴ L.H.S. = R.H.S.

Hence proved, sin 60° = 2 sin 30° cos 30°.

Prove that :

4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°) = 2

Answer

4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°) = 2

L.H.S. = 4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°)

$= 4 \times \Big(\Big(\dfrac{1}{2}\Big)^4 + \Big(\dfrac{1}{2}\Big)^4\Big) - 3 \times \Big(\Big(\dfrac{1}{\sqrt2}\Big)^2 - (1)^2\Big)\\[1em] = 4 \times \Big(\dfrac{1}{16} + \dfrac{1}{16}\Big) - 3 \times \Big(\dfrac{1}{2} - 1\Big)\\[1em] = 4 \times \Big(\dfrac{1 + 1}{16}\Big) - 3 \times \Big(\dfrac{1}{2} - \dfrac{2}{2}\Big)\\[1em] = 4 \times \Big(\dfrac{2}{16}\Big) - 3 \times \Big(\dfrac{1 - 2}{2}\Big)\\[1em] = 4 \times \Big(\dfrac{1}{8}\Big) - 3 \times \Big(\dfrac{-1}{2}\Big)\\[1em] = \dfrac{1}{2} + \dfrac{3}{2}\\[1em] = \dfrac{1 + 3}{2}\\[1em] = \dfrac{ 4}{2}\\[1em] = 2$

R.H.S. = 2

∴ L.H.S. = R.H.S.

Hence, 4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°) = 2.

If sin x = cos x and x is acute, state the value of x.

Answer

sin x = cos x

As we know that sin² x + cos² x = 1

⇒ sin² x + sin² x = 1

⇒ 2sin² x = 1

⇒ sin² x = $\dfrac{1}{2}$

⇒ sin x = $\dfrac{1}{\sqrt2}$

⇒ sin x = sin 45°

⇒ x = 45°

Hence, the value of x = 45°.

If sec A = cosec A and 0° ≤ A ≤ 90°, state the value of A.

Answer

sec A = cosec A

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

sin A = cos A

As we know that sin² A + cos² A = 1

⇒ sin² A + sin² A = 1

⇒ 2sin² A = 1

⇒ sin² A = $\dfrac{1}{2}$

⇒ sin A = $\dfrac{1}{\sqrt2}$

⇒ sin A = sin 45°

⇒ A = 45°

Hence, the value of A = 45°.

If tan θ = cot θ and 0° ≤ θ ≤ 90°, state the value of θ.

Answer

tan θ = cot θ

$\dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{\text{cos θ}}{\text{sin θ}}$

$\text{sin}^2 \text{θ} = \text{cos}^2 \text{θ}$

As we know that sin² θ + cos² θ = 1

⇒ sin² θ + sin² θ = 1

⇒ 2sin² θ = 1

⇒ sin² θ = $\dfrac{1}{2}$

⇒ sin θ = $\dfrac{1}{\sqrt2}$

⇒ sin θ = sin 45°

⇒ θ = 45°

Hence, the value of θ = 45°.

If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.

Answer

sin x = cos y

cos y = sin (90° - y)

So, sin x = sin (90° - y)

⇒ x = 90° - y

⇒ x + y = 90°

Hence, the relation between x and y is x + y = 90°.

If sin x = cos y, then x + y = 45°; write true or false.

Answer

False

Reason:

sin x = cos y

It is only possible only when x = y = 45°.

sin 45° = cos 45° = 1

sec θ. cot θ = cosec θ write true or false.

Answer

True

Reason:

sec θ. cot θ = cosec θ

L.H.S. = sec θ. cot θ

$= \dfrac{1}{\text{cos θ}} \times \dfrac{\text{cos θ}}{\text{sin θ}}\\[1em] = \dfrac{1}{\cancel{\text{cos θ}}} \times \dfrac{\cancel{\text{cos θ}}}{\text{sin θ}}\\[1em] = \dfrac{1}{\text{sin θ}}\\[1em] = \text{cosec θ}$

R.H.S. = cosec θ

∴ L.H.S. = R.H.S.

For any angle θ, state the value of :

sin² θ + cos² θ.

Answer

1

Reason:

sin² θ + cos² θ = 1

Let take θ = 30°

sin² 30° = $\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}$

cos² 30° = $\Big(\dfrac{\sqrt3}{2}\Big)^2 = \dfrac{3}{4}$

Now, sin² 30° + cos² 30°

$= \dfrac{1}{4} + \dfrac{3}{4}\\[1em] = \dfrac{1 + 3}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

State for any acute angle θ whether :

sin θ increases or decreases as θ increases.

Answer

sin θ increases as θ increases

Reason — As we know,

sin 0° = 0

sin 30° = $\dfrac{1}{2}$ = 0.5

sin 45° = $\dfrac{1}{\sqrt2}$ = 0.70

sin 60° = $\dfrac{\sqrt3}{2}$ = 0.87

sin 90° = 1

State for any acute angle θ whether :

cos θ increases or decreases as θ increases.

Answer

cos θ decreases as θ increases

Reason — As we know,

cos 0° = 1

cos 30° = $\dfrac{\sqrt3}{2}$ = 0.87

cos 45° = $\dfrac{1}{\sqrt2}$ = 0.70

cos 60° = $\dfrac{1}{2}$ = 0.5

cos 90° = 0

State for any acute angle θ whether :

tan θ increases or decreases as θ decreases.

Answer

tan θ decreases as θ decreases.

Reason — As we know,

tan 90° = not defined

tan 60° = $\sqrt3$ = 1.73

tan 45° = 1

tan 30° = $\dfrac{1}{\sqrt3}$ = 0.58

tan 0° = 0