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?

Hint:

$ \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$
$1=\sin^2 x + \cos^2 x$

Complete the square... and divide by something... then you should be able to use tangent formula
