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

$$
\frac{\cos}{1-\tan x}\cdot\frac{\cos x}{\cos x} = \frac{\cos^2 x}{\cos x-\sin x}
$$
$$
\cos x\cdot\frac{\cos x-\sin x}{\cos x-\sin x} - \frac{\cos^2 x}{\cos x-\sin x} = \frac{-\sin x\cos x}{\cos x-\sin x}
$$
