Question

Prove - $\frac{\cos 2\theta }{1+\sin 2\theta }=\tan (\frac{\pi }{4}-\theta )$

Answer 1

$LHS=\frac{\cos 2\theta }{1+\sin 2\theta }=\frac{{\cos }^2\theta -{\sin }^2\theta }{{\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta }=\frac{1-{\tan }^2\theta }{{\tan }^2\theta +1+2\tan \theta }$

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

$=\frac{\tan \left(\frac{\pi }{4}\right)-\tan \theta }{1+\tan \left(\frac{\pi }{4}\right).\tan \theta }=\tan [\frac{\pi }{4}-\theta ]=R.H.S$

Hence proved