Question

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

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $\frac{2a}{b}$
• D. $\frac{2b}{a}$

Answer 1

Answer: (D)

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