Question

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

Answer 1

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

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

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

So, $\tan (\frac{\pi }{4}+x)=1,x=n\pi .$ and $\tan x=-1,x=n\pi -\frac{\pi }{4}.$