Question

$\tan (\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b})+\tan (\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b})$ is equal to

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

Answer 1

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

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

$\Rightarrow \cos x=\frac{a}{b}$ -------------------(i)

$=\tan [\frac{\pi }{2}+\frac{x}{2}]+\tan [\frac{\pi }{2}-\frac{x}{2}]$

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

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

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

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

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

The above equation can be written as:

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

From equation(i),

$\Rightarrow \cos x=\frac{a}{b}$

We get, $=\frac{2b}{a}$