Question

Prove $\left(1+{\tan }^2\theta \right){\sin }^2\theta ={\tan }^2\theta$

Answer 1

$LHS$

$=(1+{\tan }^2\theta ){\sin }^2\theta$

$=(1+\frac{{\sin }^2\theta }{{\cos }^2\theta }){\sin }^2\theta$

$=\left(\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta }\right){\sin }^2\theta$

As we know that ${\cos }^2\theta +{\sin }^2\theta =1$

Therefore,

$=\frac{1}{{\cos }^2\theta }\times {\sin }^2\theta$

$={\left[\frac{\sin \theta }{\cos \theta }\right]}^2$

since $\frac{\sin \theta }{\cos \theta }=\tan \theta$

$LHS={\tan }^2\theta =RHS$

Hence proved.