What is $\frac{2x}{1-x^2}$ when $x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$? If
$$x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$
Find
$$\frac{2x}{1-x^2}$$
I got till here by simplification by taking the previous value of x, ie,
$$x={\frac{\sqrt{1-\cos\theta}}{1+\cos\theta}}$$
$$\frac{2\tan\theta\sqrt{1+\cos\theta}}{\cos\theta+3}$$
 A: $$x=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$
But $\cos2a=2\cos^2a-1=1-2\sin^2a$, so
$$1-\cos\theta=2\sin^2 \frac{\theta}{2}$$
$$1+\cos\theta=2\cos^2 \frac{\theta}{2}$$
Hence
$$x=\left|\tan \frac{\theta}2\right|$$
Therefore
$$\frac{2x}{1-x^2}=\frac{2\left|\tan\frac{\theta}2\right|}{1-\tan^2\frac{\theta}2}=\frac{\left|2\sin\frac{\theta}2\cos\frac{\theta}2\right|}{\cos^2\frac{\theta}2-\sin^2\frac{\theta}2}=\frac{|\sin\theta|}{\cos\theta}$$
Notice that the answer is not $\tan\theta$, when $\sin\theta<0$, that is for
$$\theta \in \bigcup_{k\in\Bbb Z} ](2k-1)\pi,2k\pi[$$
Or, if we remove also the values of $\theta$ for which $\cos\theta=0$,
$$\theta \in \bigcup_{k\in\Bbb Z} ]-\pi+2k\pi,-\pi/2+2k\pi[\;\cup\;]-\pi/2+2k\pi,2k\pi[$$
A: $$x=\sqrt\frac{1-cos(\theta)}{1+cos(\theta)}.$$ Suppose to the given equation, let it be defined such that $$y = \frac{2x}{1-x^2}$$ Just substitute x in the equation y will result $$ 1-x^2 = 1 - \frac{1-cos(\theta)}{1+cos(\theta)} \implies \frac{2 cos(\theta)}{1+cos(\theta)}.$$ similarly $$x=\sqrt\frac{1-cos(\theta)}{1+cos(\theta)} * \sqrt\frac{1+cos(\theta)}{1+cos(\theta)} \implies \sqrt\frac{sin^2(\theta)}{(1+cos(\theta))^2} $$ $$ x = \frac{sin(\theta)}{1+cos(\theta)}$$ Equation y  can be rewritten as $$y = \frac{2 sin(\theta)/(1+cos(\theta))}{2cos(\theta)/(1+cos(\theta)} $$ Hence, $$ \frac{2x}{1-x^2} = tan(\theta)$$
