Question

The value of ${\tan }^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right],\left|x\right|<\frac{1}{2},x\ne 0,$ is equal to :

• A. $\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2$
• B. $\frac{\pi }{4}+{\cos }^{-1}{x}^2$
• C. $\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}{x}^2$
• D. $\frac{\pi }{4}-{\cos }^{-1}{x}^2$

Answer 1

Answer: (A)

$x^2=cos2\theta$
$\theta =\frac{1}{2}co{s}^{-1}{x}^2$
$ta{n}^{-1}\left[\frac{\sqrt{1+cos2\theta }+\sqrt{1-cos2\theta }}{\sqrt{1+cos2\theta }-\sqrt{1-cos2\theta }}\right]$
$ta{n}^{-1}\left[\frac{1+tan\theta }{1-tan\theta }\right]$
$=ta{n}^{-1}\left[tan\left(\frac{\pi }{4}+\theta \right)\right]$
$=\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}{x}^2$