# If tanx = b/a then find the value of

Question:

If tanx = b/a then find the value ofÂ

$\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$

Solution:

According to the question,

tan x = b/a

Let,

$y=\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$

$\therefore y=\sqrt{\frac{a\left(1+\frac{b}{a}\right)}{a\left(1-\frac{b}{a}\right)}}+\sqrt{\frac{a\left(1-\frac{b}{a}\right)}{a\left(1+\frac{b}{a}\right)}}$

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

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

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

$=\frac{1+\tan x+1-\tan x}{\sqrt{1-\tan ^{2} x}}=\frac{2}{\sqrt{1-\tan ^{2} x}}$

$\therefore y=\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{2}{\sqrt{1-\tan ^{2} x}}$

$=\frac{2}{\sqrt{1-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}}$

$=\frac{2}{\frac{\sqrt{\cos ^{2} \theta-\sin ^{2} \theta}}{\cos \theta}}$

$\because \cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta$

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