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