Question

Prove $\frac{{\sin }^2(-160)}{{\sin }^{2}70}+\frac{\sin (180-\theta )}{\sin \theta }={\sec }^{2}20$

Answer 1

Formula:

$\sin (-\theta )=\sin \theta$

$\sin (180-\theta )=\sin \theta$

$\sin (90-\theta )=\cos \theta$

$\frac{{\sin }^2(-160)}{{\sin }^{2}70}+\frac{\sin (180-\theta )}{\sin \theta }$

$=\frac{{\sin }^{2}160}{{\sin }^{2}70}+\frac{\sin \theta }{\sin \theta }$

$=\frac{{\sin }^2(180-20)}{{\sin }^2(90-20)}+1$

$=\frac{{\sin }^{2}20}{{\cos }^{2}20}+1$

$={\tan }^{2}20+1$

$={\sec }^{2}20$

Hence proof