Simplify $ \frac{4\cos(x)}{1-\cos(x)} - \frac{4\sin^2(x)}{(1-\cos(x))^2}$

Question

Spent any hour trying, but can't simplify this formula.

$$ \frac{4\cos(x)}{1-\cos(x)} - \frac{4\sin^2(x)}{(1-\cos(x))^2}$$

Calculator CAS tells me the result should be $\dfrac{4}{\cos(x) - 1}$, but I can't seem to do it myself. Thanks!

Answer 1

HINT

Recall that

$$\sin^2 x=1-\cos^2 x=(1-\cos x)(1+\cos x)$$

therefore

$$\frac{4\cos x}{1-\cos x } - \frac{4\sin^2 x}{(1-\cos x)^2}=\frac{4\cos x}{1-\cos x} - \frac{4(1-\cos x)(1+\cos x)}{(1-\cos x)^2}$$

Answer 2

$\dfrac{4\cos(x)(1-\cos(x))-4\sin^2(x)}{(1-\cos(x))^2}=\dfrac{4(\cos x-1)}{(1-\cos(x))^2}=\dfrac{-4}{1-\cos(x)}=\dfrac{4}{\cos(x) - 1}$.

(Using $\sin^2x+\cos^2x=1$).