^ - 1 1 x^2 - 1 = - Vidyadip

MCQ

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

A
$\frac{\pi }{2} + {\rm{cose}}{{\rm{c}}^{ - 1}}x$
B
$\frac{\pi }{2} + {\sec ^{ - 1}}x$
✓
${\rm{cose}}{{\rm{c}}^{ - 1}}x$
D
${\sec ^{ - 1}}x$

✓

Answer

Correct option: C.

${\rm{cose}}{{\rm{c}}^{ - 1}}x$

c
(c) ${\tan ^{ - 1}}\frac{1}{{\sqrt {{x^2} - 1} }} = {\tan ^{ - 1}}\frac{1}{{\sqrt {{\rm{cose}}{{\rm{c}}^2}\theta - 1} }}$

(Putting $x = {\rm{cos}}{\rm{ec}}\,\,\theta )$

$ = {\tan ^{ - 1}}\frac{1}{{\cot \theta }} = \theta = {\rm{cose}}{{\rm{c}}^{ - 1}}x$.

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