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?
Best Answer
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.