Simplify: $\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$ Simplify:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}$$
So what I've tried was:
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{1+\cos^2x-\sin^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=$$
$$=\frac{2\cos^2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=\frac{2\cos^2x\,\tan^2x-4\cos^2(\frac{x}{2})(1-\cos x)}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{2\sin^2x-4\cos^2(\frac{x}{2})+4\cos^2(\frac{x}{2})\cos x}{(1-\cos x)(\tan^2x)}=\frac{2\cdot2\sin^2(\frac{x}{2})\cos^2(\frac{x}{2})-4\cos^2(\frac{x}{2})+4\cos^2(\frac{x}{2})\cos x}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{4\cos^2(\frac{x}{2})(\sin^2(\frac{x}{2})-1+\cos x)}{(1-\cos x)(\tan^2x)}=\frac{4\cos^2(\frac{x}{2})(\sin^2(\frac{x}{2})-1+\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=$$
$$=\frac{4\cos^2(\frac{x}{2})(-1+\cos^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=\frac{4\cos^2(\frac{x}{2})(-\sin^2(\frac{x}{2}))}{(1-\cos x)(\tan^2x)}=\frac{-2\sin^2(x)}{(1-\cos x)(\tan^2x)}$$
This is farthest I got but the answer is supposed be: $0$,
according to the solutions, I might've missed something on the process even though I checked it a couple of time..

I'm preparing for the Tel-Aviv univeristy math entry test and I took this exercise from math-exercises.com to practise (that is where the solution is from).
 A: We have that
$$\cos2x=\cos^2x-\sin^2x=2\cos^2x-1$$
$$\cos^2\left(\frac{x}{2}\right)=\frac12(\cos x +1)$$
$$\tan^2x=\frac{1-\cos^2x}{\cos^2 x}$$
thus
$$\frac{1+\cos2x}{1-\cos x}-\frac{4\cos^2(\frac{x}{2})}{\tan^2x}=$$
$$\frac{2\cos^2x}{1-\cos x}-\frac{2(\cos x +1)}{\frac{1-\cos^2x}{\cos^2 x}}=$$
$$\frac{2\cos^2x}{1-\cos x}-\frac{2\cos^2x}{1-\cos x}=0$$
A: $$u=\frac{1+cos2x}{1-cosx}-\frac{4cos^2(\frac{x}{2})}{tan^2x}$$

Using $\tan(x/2)=t$, $\sin x=\dfrac{2t}{1+t^2}$ and $\cos x=\dfrac{1-t^2}{1+t^2}$

$$U=\dfrac{2\cos^2 x}{1-cosx}-2\dfrac{(1+\cos^2x)}{\tan^2x}$$
Then
$$U=2\dfrac{(1-t^2)^2(1+t^2)}{(1+t^2)^22t^2}-2\dfrac{\frac{2}{1+t^2}}{4t^2}(1-t^2)^2$$
Eventually:
$$U=\dfrac{(1-t^2)^2}{(1+t^2)t^2}-\dfrac{(1-t^2)^2}{(1+t^2)t^2}=0$$
A: The computation can be much shorter if one uses the relevant duplication/linearisation formulæ:


*

*$\displaystyle \frac{1+\cos 2x}{1-\cos x}=\frac{2\cos^2x}{1-\cos x}=\frac{\cos^2x}{\,\smash[b]{\cfrac{1-\cos x}2}\,}=\frac{\cos^2x}{\,\sin^2\dfrac x2\,}$.


$\phantom{fg}$


*

*$\displaystyle \frac{4\cos^2\dfrac  x2}{\tan^2x}=\frac{4\cos^2\dfrac  x2\,\cos^2x}{\sin^2x}=\frac{4\cos^2\dfrac  x2\,\cos^2x}{\Bigl(2\sin\dfrac x2\cos\dfrac x2\Bigr)^2}=\frac{\cos^2x}{\,\sin^2\dfrac x2\,}$.

