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 equal to

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

Answer 1

Answer: (C)

$\frac{2}{x}$

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