Question

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 }} $
• E. $ \frac{2\tan \alpha }{\sqrt{\cos 2\alpha }} $

Answer 1

Answer: (A)

Given, $tan \, \alpha=\frac{b}{a},a>b>0 $
Now, $ \sqrt{\frac{a+b}{a-b}}-\sqrt{\frac{a-b}{a+b}} $
$=\frac{a+b-a+b}{\sqrt{a-b}.\sqrt{a+b}}=\frac{2b}{\sqrt{{{a}^{2}}-{{b}^{2}}}} $
$=\frac{2\frac{b}{a}}{\sqrt{1-{{\left( \frac{b}{a} \right)}^{2}}}} $
$=\frac{2\tan \alpha }{\sqrt{1-{{\tan }^{2}}\alpha }} $
$=\frac{2\frac{\sin \alpha }{\cos \alpha }}{\sqrt{\frac{{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }}}=\frac{2\sin a}{\sqrt{\cos 2\alpha }} $