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Mathematics · 2017
JEE Main · 8 April 2017, Shift 1 · Q89
The value of tan^-1[(√(1+x^2)+√(1-x^2))/(√(1+x^2)-√(1-x^2))], |x|<1/2, x ≠ 0, is equal to:
The value of $\displaystyle \tan ^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$, $\displaystyle |x|<\frac{1}{2}, x \neq 0$, is equal to :
Official answer
(1)
$\displaystyle \frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2$
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JEE Main 2017 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.