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 $L = P$?


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 '16 at 20:45
  • 1
    $\begingroup$ @Mathmo123 indeed, but also a term which has to be taken into account. $\endgroup$ – nippon Mar 7 '16 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. $\endgroup$ – marty cohen Mar 8 '16 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.