Answer
$\dfrac{1+\sin x}{1-\sin x}=\tan^2 (\dfrac{\pi}{4}+\dfrac{x}{2})$
Work Step by Step
Here, we have $1+\sin x=\sin (\pi/2)+\sin x$
or, $2 \sin (\dfrac{\pi/2+x}{2}) \cos (\dfrac{\pi/2 -x}{2})=2 \sin^2 (\dfrac{\pi}{4}+\dfrac{x}{2})$
Now,$1-\sin x=\sin (\pi/2) -\sin x$
or, $2 \cos (\dfrac{\pi/2+x}{2}) \sin (\dfrac{\pi/2 -x}{2})=2 \cos^2 (\dfrac{\pi}{4}+\dfrac{x}{2})$
Consider LHS: $\dfrac{1+\sin x}{1-\sin x}=\dfrac{2 \sin (\dfrac{\pi/2+x}{2}) \cos (\dfrac{\pi/2 -x}{2})}{2 \cos (\dfrac{\pi/2+x}{2}) \sin (\dfrac{\pi/2 -x}{2})}$
or, $\dfrac{2 \sin (\dfrac{\pi/2+x}{2}) \cos (\dfrac{\pi/2 -x}{2})}{2 \cos (\dfrac{\pi/2+x}{2}) \sin (\dfrac{\pi/2 -x}{2})}=\dfrac{2 \sin^2 (\dfrac{\pi}{4}+\dfrac{x}{2})}{2 \cos^2 (\dfrac{\pi}{4}+\dfrac{x}{2})}$
Thus, $\dfrac{1+\sin x}{1-\sin x}=\tan^2 (\dfrac{\pi}{4}+\dfrac{x}{2})$
Hence, the result has been proved.