$\frac{\sin ^2\left(-160^{\circ} ight)}{\sin ^2 70^{\circ}}+\frac{\sin \left(180^{\circ}- heta ight)}{\sin heta}=$

Question

$\frac{\sin ^2\left(-160^{\circ}ight)}{\sin ^2 70^{\circ}}+\frac{\sin \left(180^{\circ}-	hetaight)}{\sin 	heta}=$

Aicharya · Accepted Answer

$\begin{aligned} & \frac{\sin ^2\left(-160^{\circ}ight)}{\sin ^2 70^{\circ}}+\frac{\sin \left(180^{\circ}-	hetaight)}{\sin 	heta} \ & =\frac{\sin ^2\left(160^{\circ}ight)}{\sin ^2\left(70^{\circ}ight)}+\frac{\sin 	heta}{\sin 	heta} \ & =\frac{\sin ^2\left(90^{\circ}+70^{\circ}ight)}{\sin ^2 70^{\circ}}+1 \ & \frac{\cos ^2 70^{\circ}}{\sin ^2 70^{\circ}}+1 \ & =\cot ^2 70^{\circ}+1=\operatorname{cosec}^2 70^{\circ}=\sec ^2 20^{\circ}\end{aligned}$

$\frac{\sin ^2\left(-160^{\circ}\right)}{\sin ^2 70^{\circ}}+\frac{\sin…

Solution