If 0 < x < 1, then ^ - 1 ( 2 x^2 - 1 2x 1 - x^2 ) is equal to - Vidyadip

MCQ

If $0 < x < 1$, then ${\cot ^{ - 1}}\left( {\frac{{2{x^2} - 1}}{{2x\sqrt {1 - {x^2}} }}} \right)$ is equal to

A
$ - 2{\cot ^{ - 1}}x$
B
$\pi - 2{\cos ^{ - 1}}x$
✓
$ 2{\cos ^{ - 1}}x$
D
$2{\cos ^{ - 1}}x - \pi$

✓

Answer

Correct option: C.

$ 2{\cos ^{ - 1}}x$

c
$\cot ^{-1}\left(\frac{2 x^{2}-1}{2 x \sqrt{1-x^{2}}}\right)$

put $x = \cos \theta $ $\because $ $\theta \in \left( {0,\frac{\pi }{2}} \right)\quad \therefore \quad 0 < x < 1$

$\Rightarrow \quad \cot ^{-1}\left(\frac{\cos 2 \theta}{2 \cos \theta|\sin \theta|}\right)$

$=\cot ^{-1}(\cot 2 \theta)=2 \theta=2 \cos ^{-1} x$

$\because 2 \theta \in(0, \pi)$

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