If tan x = b/a, Find the Value of √((a+b)/(a-b)) + √((a-b)/(a+b))

Given, \[ \tan x = \frac{b}{a} \]

Using the identity, \[ \frac{1+\tan x}{1-\tan x}=\tan\left(\frac{\pi}{4}+x\right) \]

Substituting \[ \tan x=\frac{b}{a}, \] we get

\[ \frac{1+\frac{b}{a}}{1-\frac{b}{a}} = \frac{a+b}{a-b} \]

Therefore, \[ \sqrt{\frac{a+b}{a-b}} = \sqrt{\tan\left(\frac{\pi}{4}+x\right)} \]

Let \[ A=\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}} \]

Taking LCM,

\[ A= \frac{(a+b)+(a-b)} {\sqrt{(a+b)(a-b)}} \]

\[ = \frac{2a} {\sqrt{a^2-b^2}} \]

Since \[ \tan x=\frac{b}{a}, \] take \[ a=r\cos x,\qquad b=r\sin x \]

Then, \[ a^2-b^2 = r^2(\cos^2x-\sin^2x) \]

\[ = r^2\cos2x \]

Hence,

\[ A= \frac{2r\cos x}{r\sqrt{\cos2x}} \]

\[ = \frac{2\cos x}{\sqrt{\cos2x}} \]