<s>[INST] If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$. 
If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.

HINT: The answer is $-2\pm \sqrt5$.
What I have tried so far:
$$\tan{\frac{x}{2}} = \frac{1}{\sin x}-\sin x$$
$$\tan{\frac{x}{2}} = \frac{1-\sin^2 x}{\sin x}$$
$$\tan{\frac{x}{2}} = \frac{\cos^2 x}{\sin x}$$
I don't know how to solve this problem. Pls help. Thank you :)
 [/INST] Hint: Use that $$\frac{1}{\sin(x)}-\sin(x)=1/2\,{\frac {1+ \left( \tan \left( x/2 \right)  \right) ^{2}}{\tan
 \left( x/2 \right) }}-2\,{\frac {\tan \left( x/2 \right) }{1+ \left( 
\tan \left( x/2 \right)  \right) ^{2}}}
$$
</s>