Question

Prove that ${\tan }^2\theta -{\sin }^2\theta ={\tan }^2\theta {\sin }^2\theta$

Answer 1

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

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

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

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

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

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

$=$R.H.S

Hence proved.