Question

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

Answer 1

$L.H.S.$

$\frac{2\sin \theta -\sin 2\theta }{2\sin \theta +\sin 2\theta }$

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

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

$=\frac{(1-\cos \theta )}{(1+cos\theta )}$

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

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

$={\tan }^2\theta$

$R.H.S.$