Question

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

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

Answer 1

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

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

Let ${\cos }^{-1}x=\theta$

$\Rightarrow \cos \theta =x$ and $\frac{1}{2}{\cos }^{-1}x=\frac{\theta }{2}$

$=\tan (\frac{\pi }{4}+\frac{\theta }{2})+\tan (\frac{\pi }{4}-\frac{\theta }{2})$

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

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

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

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

$=2\left(\frac{1-{\tan }^2\frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}\right)$

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

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