^ -1 ( 1+x^2-1 x )= - Vidyadip

MCQ

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

A
$\tan ^{-1} x$
✓
$\frac{1}{2} \tan ^{-1} x$
C
$2 \tan ^{-1} x$
D
$3 \tan ^{-1} x$

✓

Answer

Correct option: B.

$\frac{1}{2} \tan ^{-1} x$

(B) $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$
$=\tan ^{-1}\left[\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right]$
(Putting $x=\tan \theta$ )
$=\tan ^{-1}\left[\frac{\sec \theta-1}{\tan \theta}\right]=\tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right]$
$=\tan ^{-1}\left[\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right]$
$=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)$
$=\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x$

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