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?

This is the same way to go as in the OP, maybe combining the arguments looks simpler. Let $t$ be $t=\tan(x/2)$ for the "good $x$" satisfying the given relation. Then $\displaystyle \sin x=\frac {2t}{1+t^2}$, so
$$
4=
\frac{1-\sin x}{1+\sin x}
=
\frac{(1+t^2)-2t}{(1+t^2)+2t}
=\left(\frac{1-t}{1+t}\right)^2\ .
$$
This gives for $(1-t)/(1+t)$ the values $\pm 2$, leading to the two solutions $-3$ and $-1/3$ mentioned in the OP.
