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?
 A: 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.
A: 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
A: 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.
A: $$\tan \frac{x}{2} = \frac{\sin x/2}{\cos x/2} = \frac{2 \sin x/2 \cos x/2}{2 \cos^2 x/2} = \frac{\sin x}{\cos x + 1}$$
hence with $\sin x = -\frac{3}{5}, \cos x = ±\sqrt{1 - \sin^2 x} = ±\frac{4}{5}$, $\tan \frac{x}{2} = -\frac{1}{3}, -3$.
A: See , $$\frac{1-\sin x}{1+\sin x}=4$$ $$1-\sin x=4+4\sin x$$ $$\sin x=-\frac{3}{5}$$ Hence , $x\approx-37^o$ and $\tan\frac{-37}{2}\approx-\frac{1}{3
}$
