Solve the following identity:
$$\frac{1+\tan (x)}{1-\tan (x)}=\frac{\cos (2x)}{1-\sin(2x)}$$
I can solve it by multiplying the left side by $1- \tan x$ but the method is very tedious and involves doing $(1 - \tan x)(1 - \tan x)$ and substituting $\sin$ and $\cos$ for each result.
Is there any easier method or am I just missing it completely?
Just expand the left-hand side: $$\frac{1+\tan x}{1-\tan x} = \frac{1+\frac{\sin x}{\cos x}}{1-\frac{\sin x}{\cos x}} = \frac{\cos x + \sin x}{\cos x - \sin x}.$$ Now multiply numerator and denominator by $\cos x - \sin x$: $$\frac{\cos x + \sin x}{\cos x - \sin x} = \frac{\cos^2 x - \sin^2 x}{(\cos x-\sin x )^2} = \frac{\cos 2x}{\cos^2 x + \sin^2 x - 2\sin x\cos x} = \frac{\cos 2x}{1-\sin 2x}.$$ When faced with something like this, it's never a bad idea to at least try expressing everything in terms of sine and cosine.