Question

Show that:
${\csc }^2\theta -{\tan }^2({90}^o-\theta )={\sin }^2\theta +{\sin }^2({90}^o-\theta )$

Answer 1

Since $90-\theta$ is in first quadrant, $\tan (90-\theta )=\cot \theta$ and $\sin (90-\theta )=\cos \theta$

L.H.S$={\csc }^2\theta -{\tan }^2(90-\theta )$

$={\csc }^2\theta -{\cot }^2\theta$

$=1$

$={\sin }^2\theta +{\cos }^2\theta$ since ${\sin }^2\theta +{\cos }^2\theta =1$

$={\sin }^2\theta +{\sin }^2(90-\theta )$ since $\sin (90-\theta )=\cos \theta =$R.H.S

Hence proved