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Q. The value of $\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)+\tan$ $\left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)$ is

Inverse Trigonometric Functions

Solution:

Given, $\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)+\tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{a}{b}\right)$
Let $ \frac{1}{2} \cos ^{-1} \frac{a}{b}=x \Rightarrow \cos 2 x=\frac{a}{b}$
So, $\tan \left(\frac{\pi}{4}+x\right)+\tan \left(\frac{\pi}{4}-x\right)$
$=\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x} $
$=\frac{1+\tan ^2 x+2 \tan x+1+\tan ^2 x-2 \tan x}{1-\tan ^2 x}$
$=\frac{2\left(1+\tan ^2 x\right)}{1-\tan ^2 x}=\frac{2}{\cos 2 x}=\frac{2}{\frac{a}{b}}=\frac{2 b}{a}$