Question

$ta{n}^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right]=$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

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