Question

Evaluate $\tan \left[\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right]+\tan \left[\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right]=$

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

Answer 1

The correct option is B $\frac{2b}{a}$

Let $x={\cos }^{-1}\frac{a}{b}$

$\tan [\frac{\theta }{4}+\frac{x}{2}]+\tan [\frac{\theta }{4}-\frac{x}{2}]$

$=\frac{\frac{\tan \theta }{4}+\frac{\tan x}{2}}{1-\frac{\tan \theta }{4}\times \frac{\tan x}{2}}+\frac{\frac{\tan \theta }{4}-\frac{\tan x}{2}}{1+\frac{\tan \theta }{4}\times \frac{\tan x}{2}}$

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

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

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

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

$=\frac{2}{\frac{a}{b}}$

$=\frac{2b}{a}$