# If tan x

Question:

If $\tan x=\frac{b}{a}$, then find the value of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$. [NCERT]

Solution:

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

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

$=\sqrt{\frac{1+\frac{b}{a}}{1-\frac{b}{a}}}+\sqrt{\frac{1-\frac{b}{a}}{1+\frac{b}{a}}}$

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

$=\sqrt{\frac{1+\frac{\sin x}{\cos x}}{1-\frac{\sin x}{\cos x}}}+\sqrt{\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}}$

$=\sqrt{\frac{\cos x+\sin x}{\cos x-\sin x}}+\sqrt{\frac{\cos x-\sin x}{\cos x+\sin x}}$

$=\frac{\cos x+\sin x+\cos x-\sin x}{\sqrt{(\cos x-\sin x)(\cos x+\sin x)}}$

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

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