Q. If $\tan{\alpha = \frac{b}{a},\ a > b > 0}$ and if $0 < \alpha < \frac{\pi}{4}$, then $\sqrt{\frac{a + b}{a - b}} - \sqrt{\frac{a - b}{a + b}}$ is equal to
A) $\frac{2\sin\alpha}{\sqrt{\cos{2\alpha}}}$
B) $\frac{2\cos\alpha}{\sqrt{\cos{2\alpha}}}$
C) $\frac{2\sin\alpha}{\sqrt{\sin{2\alpha}}}$
D) $\frac{2\cos\alpha}{\sqrt{\sin{2\alpha}}}$
Explanation
(a)
$\sqrt{\frac{a + b}{a - b}} - \sqrt{\frac{a - b}{a + b}}$
$= \sqrt{\frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}} - \sqrt{\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}}$
$= \sqrt{\frac{1 + \tan\alpha}{1 - \tan\alpha}} - \sqrt{\frac{1 - \tan\alpha}{1 + \tan\alpha}}$
$= \frac{\left( 1 + \tan\alpha \right) - (1 - \tan\alpha)}{\sqrt{1 - \tan^{2}\alpha}}$
$= \frac{2\tan\alpha}{\sqrt{1 - \tan^{2}\alpha}} = \frac{2\sin\alpha}{\sqrt{\cos{2\alpha}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.