Trigonometrical Ratios of Standard Angles — Chapter Test

Find the values of :

sin² 60° - cos² 45° + 3 tan² 30°

Answer

Solving,

$\Rightarrow \Big(\dfrac{\sqrt{3}}{2}\Big)^2 - \Big(\dfrac{1}{\sqrt{2}}\Big)^2 + 3 \Big(\dfrac{1}{\sqrt{3}}\Big)^2 \\[1em] \Rightarrow \dfrac{3}{4} - \dfrac{1}{2} + 3 \times \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{3 - 2}{4} + 1 \\[1em] \Rightarrow \dfrac{1}{4} + 1 \\[1em] \Rightarrow \dfrac{1 + 4}{4} \\[1em] \Rightarrow \dfrac{5}{4} \\[1em] \Rightarrow 1\dfrac{1}{4}.$

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

Find the values of :

$\dfrac{\text{2 cos}^2 45° + \text{3 tan}^2 30°}{\sqrt{3}\text{cos 30° + \text{sin 30°}}}$

Answer

Solving,

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

Hence, $\dfrac{\text{2 cos}^2 45° + \text{3 tan}^2 30°}{\sqrt{3}\text{cos 30° + \text{sin 30°}}}$ = 1.

Find the values of :

sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°

Answer

Solving,

$\Rightarrow \text{sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°} \\[1em] \Rightarrow \dfrac{2}{\sqrt{3}} \times \sqrt{3} + \text{sin 45°} \times \dfrac{1}{\text{sin 45°}} + \dfrac{\sqrt{3}}{2} \times \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow 2 + 1 + \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{4 + 2 + 1}{2} \\[1em] \Rightarrow \dfrac{7}{2} \\[1em] \Rightarrow 3\dfrac{1}{2}.$

Hence, sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60° = $3\dfrac{1}{2}.$

Taking A = 30°, verify that

(i) cos⁴ A - sin⁴ A = cos 2A

(ii) 4 cos A cos (60° - A) cos (60° + A) = cos 3A.

Answer

(i) To verify,

cos⁴ A - sin⁴ A = cos 2A

Substituting value of A in L.H.S. of the above equation, we get :

$\Rightarrow \text{cos}^4 \space A - \text{sin}^4 \space A \\[1em] \Rightarrow \text{cos}^4 \space 30° - \text{sin}^4 \space 30° \\[1em] \Rightarrow \Big(\dfrac{\sqrt{3}}{2}\Big)^4 - \Big(\dfrac{1}{2}\Big)^4 \\[1em] \Rightarrow \dfrac{9}{16} - \dfrac{1}{16} \\[1em] \Rightarrow \dfrac{8}{16} \\[1em] \Rightarrow \dfrac{1}{2}.$

Substituting value of A in R.H.S. of the above equation, we get :

⇒ cos 2A = cos 2(30°) = cos 60° = $\dfrac{1}{2}$.

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

Hence, proved that cos⁴ A - sin⁴ A = cos 2A.

(ii) To verify,

4 cos A cos (60° - A) cos (60° + A) = cos 3A.

Substituting value of A in L.H.S. of the above equation, we get :

$\Rightarrow \text{4 cos 30° cos (60° - 30°) cos (60° + 30°)} \\[1em] \Rightarrow \text{4 cos 30° cos 30° cos 90°} \\[1em] \Rightarrow 4 \times \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} \times 0 \\[1em] \Rightarrow 0.$

Substituting value of A in R.H.S. of the above equation, we get :

$\Rightarrow \text{cos 3(30°)} = \text{cos 90°} = 0.$

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

Hence, proved that 4 cos A cos (60° - A) cos (60° + A) = cos 3A

If A = 45° and B = 30°, verify that $\dfrac{\text{sin A}}{\text{cos A + sin A sin B}} = \dfrac{2}{3}$.

Answer

To verify,

$\dfrac{\text{sin A}}{\text{cos A + sin A sin B}} = \dfrac{2}{3}$.

Substituting value of A and B in L.H.S. of the equation, we get :

$\Rightarrow \dfrac{\text{sin 45°}}{\text{cos 45° + sin 45° sin 30°}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}\Big(1 + \dfrac{1}{2}\Big)} \\[1em] \Rightarrow \dfrac{\bcancel{\dfrac{1}{\sqrt{2}}}}{\bcancel{\dfrac{1}{\sqrt{2}}}\Big(\dfrac{2 + 1}{2}\Big)} \\[1em] \Rightarrow \dfrac{1}{\dfrac{3}{2}} \\[1em] \Rightarrow \dfrac{2}{3}.$

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

Hence, proved that $\dfrac{\text{sin A}}{\text{cos A + sin A sin B}} = \dfrac{2}{3}$.

Taking A = 60° and B = 30°, verify that

(i) $\dfrac{\text{sin(A + B)}}{\text{cos A cos B}} = \text{tan A + tan B}$

(ii) $\dfrac{\text{sin(A - B)}}{\text{sin A sin B}} = \text{cot B - cot A}$

Answer

(i) To verify,

$\dfrac{\text{sin(A + B)}}{\text{cos A cos B}} = \text{tan A + tan B}$.

Substituting value of A and B in L.H.S. of the equation, we get :

$\Rightarrow \dfrac{\text{sin(A + B)}}{\text{cos A cos B}} \\[1em] \Rightarrow \dfrac{\text{sin(60° + 30°)}}{\text{cos 60° cos 30°}} \\[1em] \Rightarrow \dfrac{\text{sin 90°}}{\dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}} \\[1em] \Rightarrow \dfrac{1}{\dfrac{\sqrt{3}}{4}} \\[1em] \Rightarrow \dfrac{4}{\sqrt{3}}.$

Substituting value of A and B in R.H.S. of the equation, we get :

$\Rightarrow \text{tan A + tan B} \\[1em] \Rightarrow \text{tan 60° + tan 30°} \\[1em] \Rightarrow \sqrt{3} + \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{3 + 1}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{4}{\sqrt{3}}.$

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

Hence, proved that $\dfrac{\text{sin(A + B)}}{\text{cos A cos B}} = \text{tan A + tan B}$.

(ii) To verify,

$\dfrac{\text{sin(A - B)}}{\text{sin A sin B}} = \text{cot B - cot A}$

Substituting value of A and B in L.H.S. of the equation, we get :

$\Rightarrow \dfrac{\text{sin(A - B)}}{\text{sin A sin B}} \\[1em] \Rightarrow \dfrac{\text{sin(60° - 30°)}}{\text{sin 60° sin 30°}} \\[1em] \Rightarrow \dfrac{\text{sin 30°}}{\text{sin 60° sin 30°}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2} \times \dfrac{1}{2}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{4}} \\[1em] \Rightarrow \dfrac{4}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2}{\sqrt{3}}.$

Substituting value of A and B in R.H.S. of the equation, we get :

⇒ cot B - cot A

⇒ cot 30° - cot 60°

⇒ $\sqrt{3} - \dfrac{1}{\sqrt{3}} = \dfrac{3 - 1}{\sqrt{3}}$

⇒ $\dfrac{2}{\sqrt{3}}$.

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

Hence, proved that $\dfrac{\text{sin(A - B)}}{\text{sin A sin B}} = \text{cot B - cot A}$.

If $\sqrt{2}$ tan 2θ = $\sqrt{6}$ and 0° < 2θ < 90°, find the value of

sin θ + $\sqrt{3}$ cos θ - 2 tan² θ.

Answer

Given,

$\phantom{\Rightarrow}\sqrt{2}$ tan 2θ = $\sqrt{6}$

⇒ tan 2θ = $\dfrac{\sqrt{6}}{\sqrt{2}}$

⇒ tan 2θ = $\sqrt{3}$

⇒ tan 2θ = tan 60°

⇒ 2θ = 60°

⇒ θ = $\dfrac{60°}{2}$

⇒ θ = 30°.

Substituting value of θ in sin θ + $\sqrt{3}$ cos θ - 2 tan² θ, we get :

⇒ sin 30° + $\sqrt{3}$ cos 30° - 2 tan² 30°

$\Rightarrow \dfrac{1}{2} + \sqrt{3} \times \dfrac{\sqrt{3}}{2} - 2 \times \Big(\dfrac{1}{\sqrt{3}}\Big)^2 \\[1em] \Rightarrow \dfrac{1}{2} + \dfrac{3}{2} - \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{3 + 9 - 4}{6} \\[1em] \Rightarrow \dfrac{8}{6} \\[1em] \Rightarrow \dfrac{4}{3}.$

Hence, sin θ + $\sqrt{3}$ cos θ - 2 tan² θ = $\dfrac{4}{3}.$

If 3θ is an acute angle, solve the following equation for θ :

(cosec 3θ - 2)(cot 2θ - 1) = 0.

Answer

Given,

⇒ (cosec 3θ – 2)(cot 2θ – 1) = 0

⇒ cosec 3θ – 2 = 0 or cot 2θ – 1 = 0

⇒ cosec 3θ = 2 or cot 2θ = 1

⇒ cosec 3θ = cosec 30° or cot 2θ = cot 45°

⇒ 3θ = 30° or 2θ = 45°

⇒ θ = $\dfrac{30°}{3}$ or θ = $\dfrac{45°}{2}$

⇒ θ = 10° or θ = $22\dfrac{1}{2}°$

Hence, θ = 10° or θ = $22\dfrac{1}{2}°$.

If tan (A + B) = $\sqrt{3}$, tan (A - B) = 1 and A, B (B < A) are acute angles, find the values of A and B.

Answer

Given,

⇒ tan (A + B) = $\sqrt{3}$

⇒ tan (A + B) = tan 60°

⇒ A + B = 60° ..........(1)

⇒ tan (A - B) = 1

⇒ tan (A - B) = tan 45°

⇒ A - B = 45° ..........(2)

Adding (1) and (2) we get :

⇒ A + B + A - B = 60° + 45°

⇒ 2A = 105°

⇒ A = $\dfrac{105}{2}$

⇒ A = $52\dfrac{1}{2}\degree$

Substituting value of A in (2) we get :

⇒ $\dfrac{105}{2}$ - B = 45°

⇒ B = $\dfrac{105}{2}$ - 45°

⇒ B = $\dfrac{105 - 90}{2}$

⇒ B = $\dfrac{15}{2}$

⇒ B = $7\dfrac{1}{2}\degree$

Hence, A = $52\dfrac{1}{2}\degree$ and B = $7\dfrac{1}{2}\degree$

Without using trigonometrical tables, evaluate the following :

sin² 28° + sin² 62° - tan² 45°

Answer

Solving,

⇒ sin² 28° + sin² 62° - tan² 45°

⇒ sin² 28° + sin² (90° - 28°) - tan² 45°

⇒ sin² 28° + cos² 28° - 1 [∵ sin (90 - θ) = cos θ]

⇒ 1 - 1 [∵ sin² θ + cos² θ]

⇒ 0.

Hence, sin² 28° + sin² 62° - tan² 45° = 0.

Without using trigonometrical tables, evaluate the following :

$\dfrac{\text{2 cos 27°}}{\text{sin 63°}} + \dfrac{\text{tan 27°}}{\text{cot 63°}} + \text{cos 0°}$

Answer

Solving,

$\Rightarrow \dfrac{\text{2 cos 27°}}{\text{sin 63°}} + \dfrac{\text{tan 27°}}{\text{cot 63°}} + \text{cos 0°} \\[1em] \Rightarrow \dfrac{\text{2 cos 27°}}{\text{sin (90° - 27°)}} + \dfrac{\text{tan 27°}}{\text{cot (90° - 27°)}} + 1 \\[1em] \Rightarrow \dfrac{\text{2 cos 27°}}{\text{cos 27°}} + \dfrac{\text{tan 27°}}{\text{tan 27°}} + 1 \\[1em] \Rightarrow 2 + 1 + 1 \\[1em] \Rightarrow 4.$

Hence, $\dfrac{\text{2 cos 27°}}{\text{sin 63°}} + \dfrac{\text{tan 27°}}{\text{cot 63°}} + \text{cos 0°}$ = 4.

Without using trigonometrical tables, evaluate the following :

cos 18° sin 72° + sin 18° cos 72°

Answer

Solving,

⇒ cos 18° sin 72° + sin 18° cos 72°

[As, sin (90 - θ) = cos θ and cos (90 - θ) = sin θ]

⇒ cos 18° sin (90° - 18°) + sin 18° cos (90° - 18°)

⇒ cos 18° cos 18° + sin 18° sin 18°

⇒ sin² 18° + cos² 18°

⇒ 1.

Hence, cos 18° sin 72° + sin 18° cos 72° = 1.

Without using trigonometrical tables, evaluate the following :

5 sin 50° sec 40°- 3 cos 59° cosec 31°

Answer

Solving,

⇒ 5 sin 50° sec 40°- 3 cos 59° cosec 31°

⇒ 5 sin 50° sec (90° - 50°) - 3 cos 59° cosec (90° - 59°)

[As, sec (90 - θ) = cosec θ and cosec (90 - θ) = sec θ]

⇒ 5 sin 50° cosec 50° - 3 cos 59° sec 59°

⇒ $\text{5 sin 50°} \times \dfrac{1}{\text{sin 50°}} - \text{3 cos 59°} \times \dfrac{1}{\text{cos 59°}}$

⇒ 5 - 3

⇒ 2.

Hence, 5 sin 50° sec 40°- 3 cos 59° cosec 31° = 2.

Prove that :

$\dfrac{\text{cos (90° - θ) sec (90° - θ) tan θ}}{\text{cosec (90 - θ) sin (90 - θ) cot (90 - θ)}} + \dfrac{\text{tan (90 - θ)}}{\text{cot θ}} = 2.$

Answer

Solving L.H.S. of above equation,

$\Rightarrow \dfrac{\text{cos (90° - θ) sec (90° - θ) tan θ}}{\text{cosec (90 - θ) sin (90 - θ) cot (90 - θ)}} + \dfrac{\text{tan (90 - θ)}}{\text{cot θ}} \\[1em] \Rightarrow \dfrac{\text{sin θ. cosec θ. tan θ}}{\text{sec θ. cos θ. tan θ}} + \dfrac{\text{cot θ}}{\text{cot θ}} \\[1em] \Rightarrow \dfrac{\text{sin θ } \times \dfrac{1}{\text{sin θ}}\times \text{ tan θ}}{\dfrac{1}{\text{cos θ}} \times \text{. cos θ. tan θ}} + \dfrac{\text{cot θ}}{\text{cot θ}} \\[1em] \Rightarrow 1 + 1 \\[1em] \Rightarrow 2.$

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

Hence, proved that $\dfrac{\text{cos (90° - θ) sec (90° - θ) tan θ}}{\text{cosec (90 - θ) sin (90 - θ) cot (90 - θ)}} + \dfrac{\text{tan (90 - θ)}}{\text{cot θ}} = 2.$

When 0° < A < 90°, solve the following equations :

(i) sin 3A = cos 2A

(ii) tan 5A = cot A

Answer

(i) Given,

⇒ sin 3A = cos 2A

⇒ sin 3A = sin (90° - 2A)

⇒ 3A = 90° - 2A

⇒ 5A = 90°

⇒ A = $\dfrac{90}{5}$

⇒ A = 18°.

Hence, A = 18°.

(ii) Given,

⇒ tan 5A = cot A

⇒ tan 5A = tan (90° - A)

⇒ 5A = 90° - A

⇒ 6A = 90°

⇒ A = $\dfrac{90}{6}$

⇒ A = 15°.

Hence, A = 15°.

Find the value of θ if

(i) sin (θ + 36°) = cos θ, where θ and θ + 36° are acute angles.

(ii) sec 4θ = cosec (θ - 20°), where 4θ and θ - 20° are acute angles.

Answer

(i) Given,

⇒ sin (θ + 36°) = cos θ

⇒ sin (θ + 36°) = sin (90° - θ)

⇒ θ + 36° = 90° - θ

⇒ 2θ = 90° - 36°

⇒ 2θ = 54°

⇒ θ = $\dfrac{54}{2}$

⇒ θ = 27°.

Hence, θ = 27°.

(ii) Given,

⇒ sec 4θ = cosec (θ - 20°)

⇒ sec 4θ = sec [90° - (θ - 20°)]

⇒ 4θ = [90° - (θ - 20°)]

⇒ 4θ = 110° - θ

⇒ 5θ = 110°

⇒ θ = $\dfrac{110}{5}$

⇒ θ = 22°.

Hence, θ = 22°.

In the adjoining figure, ABC is right-angled triangle at B and ABD is right angled triangle at A. If BD ⊥ AC and BC = $2\sqrt{3}$ cm, find the length of AD.

Answer

In △ABC,

tan 30° = $\dfrac{\text{Perpendicular}}{\text{Base}}$

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{BC}{AB} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{2\sqrt{3}}{AB} \\[1em] \Rightarrow AB = 6 \text{ cm}.$

In △ABE,

90° = 30° + ∠ABE [As, exterior angle is equal to sum of two opposite interior angles]

∠ABE = 90° - 30° = 60°.

From figure,

∠ABD = ∠ABE = 60°

In △ABD,

tan ∠ABD = tan 60° = $\dfrac{\text{Perpendicular}}{\text{Base}}$

$\Rightarrow \sqrt{3} = \dfrac{AD}{AB} \\[1em] \Rightarrow AD = AB\sqrt{3} \\[1em] \Rightarrow AD = 6\sqrt{3} \text{ cm}.$

Hence, AD = $6\sqrt{3}$ cm.