Question:
The value of $\sin ^{2} 29^{\circ}+\sin ^{2} 61^{\circ}$ is
(a) 1
(b) 0
(c) $2 \sin ^{2} 29^{\circ}$
(d) $2 \cos ^{2} 61^{\circ}$
Solution:
The given expression is $\sin ^{2} 29^{\circ}+\sin ^{2} 61^{\circ}$.
$\sin ^{2} 29^{\circ}+\sin ^{2} 61^{\circ}$
$=\sin ^{2} 29^{\circ}+\left(\sin 61^{\circ}\right)^{2}$
$=\sin ^{2} 29^{\circ}+\left\{\sin \left(90^{\circ}-29^{\circ}\right)\right\}^{2}$
$=\sin ^{2} 29^{\circ}+\left(\cos 29^{\circ}\right)^{2}$
$=\sin ^{2} 29^{\circ}+\cos ^{2} 29^{\circ}$
$=1$
Hence, the correct option is (a).