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
(b) Given expression, $\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}$
$=\frac{\sin ^{2} 22^{\circ}+\sin ^{2}\left(90^{\circ}-22^{\circ}\right)}{\cos ^{2}\left(90^{\circ}-68^{\circ}\right)+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin \left(90^{\circ}-63^{\circ}\right)$
$=\frac{\sin ^{2} 22^{\circ}+\cos ^{2} 22^{\circ}}{\sin ^{2} 68^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \cdot \cos 63^{\circ}\left[\begin{array}{ll}\because & \sin \left(90^{\circ}-\theta\right)=\cos \theta \\ \text { and } \cos \left(90^{\circ}-\theta\right)=\sin \theta\end{array}\right]$
$=\frac{1}{1}+\left(\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}\right)$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$=1+1=2$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.