# If tan x

Question:

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

Solution:

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

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

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

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

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

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

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