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 $L = P$?


3 Answers 3


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 $-2\tan\alpha$.


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.

  • 3
    $\begingroup$ An extra square root term, yes. But a very manageable one. Upvoters, please consider that the second line of this answer is discouraging what is probably the easiest approach to the problem. $\endgroup$
    – Mathmo123
    Mar 7, 2016 at 20:45
  • 1
    $\begingroup$ @Mathmo123 indeed, but also a term which has to be taken into account. $\endgroup$
    – nippon
    Mar 7, 2016 at 20:46

(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.

  • 1
    $\begingroup$ Fixed now. Thanks. $\endgroup$ Mar 8, 2016 at 21:37

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