Question

prove that 2tan^-1x = tan^2x/1-x^2

Answer 1

$\tan {}^{ - 1} a + {tan}^{ - 1} b =arc \tan( \frac{a + b}{1 - ab} ) \\ put \: a = b = x \\ 2tan {}^{ - 1} x = arc\tan( \frac{2x}{1 - {x}^{2} } )$

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Answer 2

Hey friend, Harish here.

Here is your answer.

To Prove:

$2 \tan ^{-1} x = \tan^{-1} \bigl( \frac{2x}{1-x^2} \bigr )$

Solution:

$let \ \tan^{-1} x = \theta \\ \\ Then \ \ , \tan \theta = x \\ \\ We\ know\ that , \\ \\ \tan 2\theta =\bigl ( \frac{2 \tan \theta}{1- \tan^2 \theta} \bigr ) \\ \\ 2\theta = \tan^{-1} \bigl ( \frac{2\tan \theta}{1- \tan^2 \theta} \bigr ) \\ \\ Now, substitute \ the\ value\ of \ \tan \theta \ and \ \theta . \\ \\ Then, \\ \\ 2\tan^{-1}x = \tan^{-1} \bigl ( \frac{2x}{1-x^2} \bigr )$

Hence proved.
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Hope my answer is helpful to you.