Question

Prove that:

${\tan }^2\theta -{\sin }^2\theta ={\tan }^2\theta .{\sin }^2\theta$.

Answer 1

Now, ${\tan }^2\theta -{\sin }^2\theta$

$=\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta$

$=\left({\sin }^2\theta \right)\left[\frac{1-{\cos }^2\theta }{{\cos }^2\theta }\right]$

$=\frac{{\sin }^2\theta .{\sin }^2\theta }{{\cos }^2\theta }$ $[\because 1-{\cos }^2\theta ={\sin }^2=\theta ]$

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