Question

$\tan (\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}x)+\tan (\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x)$ , $x\ne 0$ is equalt to-

• A. $x$
• B. $2x$
• C. $\frac{2}{x}$
• D. None of these

Answer 1

The correct option is C $\frac{2}{x}$

Consider, $\tan [\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}x]+\tan [\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x]$
Substitute ${\cos }^{-1}x=\theta$
$\tan [\frac{\pi }{4}+\frac{\theta }{2}]+\tan [\frac{\pi }{4}-\frac{\theta }{2}]$
$=\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}+\frac{1-\tan \frac{\theta }{2}}{1+\tan \frac{\theta }{2}}$
$=\frac{2(1+{\tan }^2\frac{\theta }{2})}{1-{\tan }^2\frac{\theta }{2}}$
$=\frac{2}{\cos \theta }$
$=\frac{2}{x}$