# Solve this

Question:

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

Solution:

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

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

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

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

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

$=\left\{1+\sin ^{2} 63^{\circ}+\cos ^{2} 63^{\circ}\right\}$

$=\{1+1\} \quad$ (using the identity: $\sin ^{2} \theta+\cos ^{2} \theta=1$ )

$=2$

Hence, $\left\{\frac{\sin ^{2} 22^{\circ}+\sin ^{2} 68^{\circ}}{\cos ^{2} 22^{\circ}+\cos ^{2} 68^{\circ}}+\sin ^{2} 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}\right\}=2$.