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?

You have\begin{align}\frac{1-\sin(x)}{1+\sin(x)}=4&\iff\frac{\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)-2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)+2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}=4\\&\iff\left(\frac{\sin\left(\frac x2\right)-\cos\left(\frac x2\right)}{\sin\left(\frac x2\right)+\cos\left(\frac x2\right)}\right)^2=4\\&\iff\frac{\sin\left(\frac x2\right)-\cos\left(\frac x2\right)}{\sin\left(\frac x2\right)+\cos\left(\frac x2\right)}=\pm2\\&\iff\frac{\tan\left(\frac x2\right)-1}{\tan\left(\frac x2\right)+1}=\pm2\end{align}The only solution of the equation $\frac{\tan\left(\frac x2\right)-1}{\tan\left(\frac x2\right)+1}=2$ is $\tan\left(\frac x2\right)=-3$, which is on that list, whereas the only solution of the equation $\frac{\tan\left(\frac x2\right)-1}{\tan\left(\frac x2\right)+1}=-2$ is $\tan\left(\frac x2\right)=-\frac13$, which is not on that list.
So, the problem has two solutions, but only one of them is on the list of options.
