Question

Find the value of $\tan \left[\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}\right]+\tan \lfloor \frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}\rfloor$.

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

Answer 1

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

Given, $\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b}]+\tan [\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b}]$

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

$⟹\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}$

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

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

We know,

$\cos 2x=\frac{(1-ta{n}^{2}x)}{(1+ta{n}^{2}x)}=\frac{a}{b}=2\times \frac{b}{a}=\frac{2b}{a}.$