Question

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

• A. $\frac{1}{x}$
• B. $x$
• C. $\frac{2}{x}$
• D. $2x$

Answer 1

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

We have, $\tan (\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}x)+\tan (\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x)$

$let{\cos }^{-1}x=t\Rightarrow x=\cos t$

$\Rightarrow$ $\tan (\frac{x}{4}+\frac{t}{2})+\tan (\frac{x}{4}-\frac{t}{2})$

$=\frac{\tan \left(\frac{\pi }{4}\right)+\tan \left(\frac{t}{2}\right)}{1-\tan \left(\frac{\pi }{4}\right)\tan \left(\frac{t}{2}\right)}+\frac{\tan \left(\frac{\pi }{4}\right)-\tan \left(\frac{t}{2}\right)}{1+\tan \left(\frac{\pi }{4}\right)\tan \left(\frac{t}{2}\right)}$

$=\frac{1+\tan \left(\frac{t}{2}\right)}{1-\tan \left(\frac{t}{2}\right)}+\frac{1-\tan \left(\frac{t}{2}\right)}{1+\tan \left(\frac{t}{2}\right)}$

$=\frac{{[1+\tan \left(\frac{t}{2}\right)]}^2+{[1-\tan \left(\frac{t}{2}\right)]}^2}{(1-\tan \left(\frac{t}{2}\right))(1+\tan \left(\frac{t}{2}\right))}$

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

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

$=\frac{2}{x}$