Verifying the trigonometric identity $\cos{x} - \frac{\cos{x}}{1 - \tan{x}} = \frac{\sin{x} \cos{x}}{\sin{x} - \cos{x}}$ I have the following trigonometric identity
$$\cos{x} - \frac{\cos{x}}{1 - \tan{x}} = \frac{\sin{x} \cos{x}}{\sin{x} - \cos{x}}$$
I've been trying to verify it for almost 20 minutes but coming up with nothing
Thank you

Observe that we need to eliminate $\tan x$
So, using $\tan x=\frac{\sin x}{\cos x},$
$$\cos x-\frac{\cos x}{1-\tan x}$$
$$=\cos x\left(1-\frac{1}{1-\frac{\sin x}{\cos x}}\right)$$
$$=\cos x\left(1-\frac{\cos x}{\cos x-\sin x}\right) (\text{ multiplying numerator & denominator by }\cos x)$$
$$=\cos x\left(1+\frac{\cos x}{\sin x-\cos x}\right)$$
$$=\cos x\left(\frac{\sin x}{\sin x-\cos x}\right)$$
