NCERT Exercise 8.2

1

Evaluate the following:

(i) $sin 60° cos30° + sin30° cos60°$

(ii) $2tan^{2}45° + cos^{2}30° - sin^{2}60°$

(iii) $\dfrac{cos45°}{sec30° + cosec60°}$

(iv) $\dfrac{sin30° + tan45° - cosec60°}{sec30° + cos60° + cot45°}$

(v) $\dfrac{5cos^{2}60°+4sec^{2}30°-tan^{2}45°}{sin^{2}30 + cos^{2}30°}$

Solution

(i) $sin 60° cos30° + sin30° cos60°$

$=\left ( \dfrac{1}{2} \right )\left ( \dfrac{1}{2} \right )+\left ( \dfrac{\sqrt{3}}{2} \right )\left ( \dfrac{\sqrt{3}}{2} \right )$ $=\dfrac{1}{4}+\dfrac{3}{4}=1$

(ii) $2tan^{2}45° + cos^{2}30° − sin^{2}60°$

$=2(1)^{2}+\left ( \dfrac{\sqrt{3}}{2} \right )^{2}-\left ( \dfrac{\sqrt{3}}{2} \right )^{2}$ $=2+\dfrac{3}{4}-\dfrac{3}{4}=2$

(iii) $\dfrac{cos45°}{sec30° + cosec60°}$

$=\dfrac{\dfrac{1}{\sqrt{2}}}{ \dfrac{2}{\sqrt{3}}+2}=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}} }$

$=\dfrac{\sqrt{3}}{\sqrt{2}(2+2\sqrt{3})}=\dfrac{\sqrt{3}}{(2\sqrt{2}+2\sqrt{6})}$

$=\dfrac{\sqrt{3}(2\sqrt{6}-2\sqrt{2})}{\left ( 2\sqrt{6}+2\sqrt{2} \right )\left ( 2\sqrt{6}-2\sqrt{2} \right )}$

$=\dfrac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{ \left ( 2\sqrt{6} \right )^{2}-\left ( 2\sqrt{2} \right )^{2}}=\dfrac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{ 24-8}$

$=\dfrac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{ 16}=\dfrac{\sqrt{18}-\sqrt{6}}{8}=\dfrac{3\sqrt{2}-\sqrt{6}}{8}$



(iv) $\dfrac{sin30° + tan45° - cosec60°}{sec30° + cos60° + cot45°}$

$= \dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{ \dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1}$ $=\dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{3}{2}+\dfrac{2}{\sqrt{3}}}$ $=\dfrac{\dfrac{3\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{3\sqrt{3}+4}{2\sqrt{3}}}=\dfrac{(3\sqrt{3}-4)}{(3\sqrt{3}+4)}$ $=\dfrac{\left ( 3\sqrt{3}-4 \right )\left ( 3\sqrt{3}-4 \right )}{\left ( 3\sqrt{3}-4 \right )\left ( 3\sqrt{3}+4 \right )}$ $=\dfrac{\left ( 3\sqrt{3}-4 \right )^{2}}{\left ( 3\sqrt{3} \right )^{2}-(4)^{2}}$ $=\dfrac{43-24\sqrt{3}}{11}$

(v) $\dfrac{5cos^{2}60°+4sec^{2}30°-tan^{2}45°}{sin^{2}30 + cos^{2}30°}$

$= \dfrac{5\left ( \dfrac{1}{2} \right )^{2}+4\left ( \dfrac{2}{\sqrt{3}} \right )^{2}-(1)^{2}}{ \left ( \dfrac{1}{2} \right )^{2}+\left ( \dfrac{\sqrt{}3}{2} \right )^{2}}$ $=\dfrac{5\left ( \dfrac{1}{4} \right )+\left ( \dfrac{16}{3} \right )-1 }{\left ( \dfrac{1}{2} \right )^{2}+\left ( \dfrac{\sqrt{3}}{2} \right )^{2}}$ $=\dfrac{5\left ( \dfrac{1}{4} \right )+\left ( \dfrac{16}{3} \right )-1}{\dfrac{1}{4}+\dfrac{3}{4}}$ $=\dfrac{67}{12}$

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2

Choose the correct option and justify your choice:

(i) $\dfrac{2 tan 30°}{1+tan^{2} 30°}$ =

(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°

(ii) $\dfrac{1-tan^{2}45°}{1+tan^{2}45°}=$

(A) tan 90° (B) 1 (C) sin 45° (D) 0

(iii) sin 2A = 2 sin A is true when A =

(A) 0° (B) 30° (C) 45° (D) 60°

(iv) $\dfrac{2 tan30°}{1-tan^{2}30°}=$

(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

Solution
(i) $\dfrac{2 tan 30°}{1+tan^{2} 30°}$ =

$= \dfrac{2 ×\dfrac{1}{\sqrt{3}}}{1 + \left(\dfrac{1}{\sqrt{3}}\right)^2}$ $= \dfrac{\dfrac{2}{\sqrt{3}}}{1 + \dfrac{1}{3}}$

$= \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}$ $= \dfrac{6}{4\sqrt{3}}$ $= \dfrac{\sqrt{3}}{2}$

As sin 60° $= \dfrac{\sqrt{3}}{2}$, option (A) is correct.

(ii) $\dfrac{1-tan^{2}45°}{1+tan^{2}45°}=$

$\dfrac{1-1^{2}}{1+1^{2}}=$ $\dfrac{1-1}{1+1}=$ $0$

Hence option (D) is correct.

(iii) For A = 0° sin 2A = sin 0° = 0

2 sinA = 2sin 0° = 2(0) = 0

Hence, option (A) is correct.

(iv) $\dfrac{2 tan30°}{1-tan^{2}30°}=$

$= \dfrac{2 ×\dfrac{1}{\sqrt{3}}}{1 - \left(\dfrac{1}{\sqrt{3}}\right)^2}$ $= \dfrac{\dfrac{2}{\sqrt{3}}}{1 - \dfrac{1}{3}}$

$= \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}}$ $= \dfrac{6}{2\sqrt{3}}$ $= \sqrt{3}$

As tan 60° = $= \sqrt{3}$, option (C) is correct.

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3

If tan(A+B)=$\sqrt{3}$ and tan(A-B)=$\dfrac{1}{{\sqrt{3}}}$; 0° < A+B ≤ 90°; A > B, find A and B.

Solution
tan(A + B) = $\sqrt{3}$
$⇒$tan(A + B) = tan 60°
$⇒$ A + B = 60°

tan(A - B) = $\dfrac{1}{\sqrt{3}}$
$⇒$tan(A - B) = tan 30°
$⇒$ A - B = 30°

Adding both the equations, we get 2A = 90° ⇒ A = 45°

As A + B = 60, B = 15°
∴ ∠A = 45° and ∠B = 15°

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4

State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.

(i) sin (A + B) = sin A + sin B Let A = 30° and B = 60°
sin (A + B) = sin (30° + 60°)
= sin 90° = 1 sin A + sin B = sin 30° + sin 60°
$=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}=\dfrac{1+\sqrt{3}}{2}$
As the values do not match, sin (A + B) ≠ sin A + sin B Hence, the given statement is false.


(ii) The value of sin θ increases as θ increases.
sin 0° = 0 sin 30° =$\dfrac{1}{2}$, sin 45°=$\dfrac{1}{\sqrt{2}}$, sin 60°=$\dfrac{\sqrt{3}}{2}$ and sin 90° = 1
Hence, the given statement is true in the interval of 0° < θ < 90°


(iii) The value of cos θ increases as θ increases. cos 30°=$\dfrac{\sqrt{3}}{2}$, cos 45°=$\dfrac{1}{\sqrt{2}}$, cos 60°=$\frac{1}{2}$ and cos 90° = 0
Hence, the given statement is false in the interval of 0° < θ < 90°


(iv) sin θ = cos θ for all values of θ.
This is true when θ = 45° and not for other values of θ
Hence, the given statement is false.


(v) cot A is not defined for A = 0° cot A = $\dfrac{cos A}{sin A}$
For A = 0°, cos A = 1 and sin A = 0
cot A = $\dfrac{cos 0°}{sin 0°}$ = undefined
Hence, the given statement is true.