<s>[INST] If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac {x}2$? 
If $\dfrac{1-\sin x}{1+\sin x}=4$, what is the value of $\tan \frac
{x}2$?
$1)-3\qquad\qquad2)2\qquad\qquad3)-2\qquad\qquad4)3$

Here is my method:
$$1-\sin x=4+4\sin x\quad\Rightarrow\sin x=-\frac{3}5$$
We have $\quad\sin x=\dfrac{2\tan(\frac x2)}{1+\tan^2(\frac{x}2)}=-\frac35\quad$. by testing the options we can find out $\tan(\frac x2)=-3$ works (although by solving the quadratic I get $\tan(\frac x2)=-\frac13$ too. $-3$ isn't the only possible value.)
I wonder is it possible to solve the question with other (quick) approaches?
 [/INST] $$\tan \frac{x}{2} = \frac{\sin x/2}{\cos x/2} = \frac{2 \sin x/2 \cos x/2}{2 \cos^2 x/2} = \frac{\sin x}{\cos x + 1}$$
hence with $\sin x = -\frac{3}{5}, \cos x = ±\sqrt{1 - \sin^2 x} = ±\frac{4}{5}$, $\tan \frac{x}{2} = -\frac{1}{3}, -3$.
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