Question

If $\tan x=\frac{b}{a}$ find $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$

Answer 1

Given, $\tan x=\frac{b}{a}$

$\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=(\sqrt{\frac{1+\tan x}{1-\tan x}}+\sqrt{\frac{1-\tan x}{1+\tan x}})$

$=(\sqrt{\frac{1+\tan x}{1-\tan x}}+\sqrt{\frac{1-\tan x}{1+\tan x}})$

$=\left(\frac{(1+\tan x)+(1-\tan x)}{\sqrt{1-{\tan }^{2}x}}\right)$

$=\left(\frac{2}{\sqrt{1-{\tan }^{2}x}}\right)$

$=\left(\frac{2\cos x}{\sqrt{{\cos }^{2}x-{\sin }^{2}x}}\right)$

$=\frac{2\cos x}{\sqrt{\cos x}}$