Question

Prove that:
$se{c}^2\theta -co{t}^2({90}^{\circ }-\theta )=co{s}^2({90}^{\circ }-\theta )+co{s}^2\theta .$

Answer 1

To prove : $se{c}^2\theta -co{t}^2({90}^{\circ }-\theta )=co{s}^2({90}^{\circ }-\theta )+co{s}^2\theta .$
$L.H.S.=se{c}^2\theta -co{t}^2({90}^{\circ }-\theta )$
$=se{c}^2\theta -(tan\theta {)}^2$
$=se{c}^2\theta -tan{\theta }^2$
$=1$
$R.H.S.=co{s}^2(90-\theta )+co{s}^2\theta$
$=\left[co{s}^2\right(90-\theta \left)\right]+co{s}^2\theta$
$=si{n}^2\theta +co{s}^2\theta$
$=si{n}^2\theta +co{s}^2\theta$
$=1$
Hence, L.H.S. = R.H.S. Hence Proved