Offline paper
Mathematics · 2015
JEE Main 2015 · 2015-04-04 Offline · Q89
Let tan^-1 y= tan^-1 x+ tan^-1((2 x)/(1-x^2)), where |x|<1/(√ 3). Then a value of y is:
Let
$$\tan ^{-1} y=\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^2}\right),
$$
where $\displaystyle |x|<\frac{1}{\sqrt{3}}$. Then a value of $\displaystyle y$ is :
Official answer
(1)
$\displaystyle \frac{3 x-x^3}{1-3 x^2}$
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JEE Main 2015 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.