Does $L^2 = P^2$ implies that $L = P$? I have encountered a problem when solving this problem:
Assume that $\alpha \in (\pi, \frac{3}{2} \pi)$, then prove $\sqrt{\frac{1 + \sin \alpha}{1 - \sin \alpha}} - \sqrt{\frac{1 - \sin \alpha}{1 + \sin \alpha}} = -2 \tan \alpha$
The most popular way to solve this kind of problems is to take left-hand side of the equation and prove that it is equal to the right-hand side. But this time, it is not so easy, because I can't see any way to transform LHS into RHS. My question is - can I take the square of LHS and prove that it equals to the RHS? 
In the language of mathematics: does $L^2 = P^2$ implies that $L = P$?
 A: No. Knowing that $L^2 = P^2$ only tells us that either $L =  P$ or $L=-P$.
However, that doesn't mean that this is not the correct approach. If you can show that $L^2 = P^2$, then you have still narrowed down the LHS to two possible values: suppose you now that 
$$LHS^2 = (-2\tan\alpha)^2.$$
This means that either the LHS is $2\tan\alpha$ or it is $-2\tan\alpha$. Here, you can use the condition $\alpha \in (\pi, \frac{3}{2} \pi)$ to show that the LHS must indeed be $-2\tan\alpha$.
A: No.
$L^2 = P^2$ implies $L= P$ OR $L = - P$. 
Also note that squaring the LHS of your equation will leave an extra square root term. 
A: (I made it a little more general)
Note that
$\sqrt{x}-\sqrt{\frac1{x}}
=\sqrt{x}-\frac1{\sqrt{x}}
=\frac{x-1}{\sqrt{x}}
$.
If
$x = \frac{1+y}{1-y}
$,
this is
$\begin{array}\\
\frac{x-1}{\sqrt{x}}
&=\frac{\frac{1+y}{1-y}-1}{\sqrt{\frac{1+y}{1-y}}}\\
&=\frac{1+y-(1-y)}{(1-y)\sqrt{\frac{1+y}{1-y}}}\\
&=\frac{2y}{\sqrt{(1+y)(1-y)}}\\
&=\frac{2y}{\sqrt{1-y^2}}\\
\end{array}
$
In your case,
with
$y = \sin a$,
this becomes
$\frac{2\sin a}{\sqrt{1-\sin^2a}}
=\pm\frac{2\sin a}{\cos a}
=\pm 2\tan a
$.
The restriction of $a$
then decides the sign.
