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