Question

Prove that ${2tan}^{-1}x={cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

Answer 1

b'

Let ${\tan }^{-1}x=theta\Rightarrow \tan \theta =x$

We know that $\cos 2\theta =\frac{1-{\tan }^{2}theta}{1+{\tan }^2\theta }$

$\Rightarrow \cos 2\theta =\frac{1-x^2}{1+x^2}$ where $x=\tan \theta$

$\Rightarrow 2\theta ={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$\Rightarrow 2{\tan }^{-1}x={cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ where $\theta ={\tan }^{-1}x$

Hence proved.

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