Question

Prove the following identities.
$\frac{(1+cosx)(1+cos2x)}{(1+sinx)(1-cos2x)}=\frac{1-sinx}{1-cosx}$

Answer 1

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

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

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

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

$=\frac{1-\sin x}{1-\cos x}=RHS$