Question

Solve: $\frac{1}{\tan x}-1=\frac{\cos 2x}{1+\tan x}.$

Answer 1

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

$\sin 2x=2$

So, no value esist but, $\tan x=\pm 1,x=n\pi \pm \frac{\pi }{4}.$