Question

Prove that :
$cose{c}^2({90}^{\circ }-\theta )-ta{n}^2\theta =co{s}^2({90}^{\circ }-\theta )+co{s}^2\theta$

Answer 1

We have

LHS = $cose{c}^2({90}^{\circ }-\theta )-ta{n}^2\theta$

$[\because cose{c}^2({90}^{\circ }-\theta )=sec\theta ]$

$\Rightarrow LHS=se{c}^2\theta -ta{n}^2\theta =1$

$[\because se{c}^2\theta -ta{n}^2\theta =1]$

and

RHS = $co{s}^2({90}^{\circ }-\theta )+co{s}^2\theta$

$\Rightarrow RHS=si{n}^2\theta +co{s}^2\theta =1$

$[\because cos({90}^{\circ }-\theta )=sin\theta ]$

Hence , LHS = RHS