# The value of the expression

Question:

The value of the expression

$\left(\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right)$ is

(a) 3

(b) 2

(c) 1

(d) 0

Solution:

(b) Given expression, $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}$

$=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}\left(90^{\circ}-22^{\circ}\right)}{\cos ^{2}\left(90^{\circ}-68^{\circ}\right)+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin \left(90^{\circ}-63^{\circ}\right)$

$=\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\sin ^{2} 68^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \cdot \cos 63^{\circ}\left[\begin{array}{ll}\because & \sin \left(90^{\circ}-\theta\right)=\cos \theta \\ \text { and } \cos \left(90^{\circ}-\theta\right)=\sin \theta\end{array}\right]$

$=\frac{1}{1}+\left(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}\right)$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$=1+1=2$