( ^ - 1 x) = - Vidyadip

MCQ

$\cos ({\tan ^{ - 1}}x) = $

A
$\sqrt {1 + {x^2}} $
✓
$\frac{1}{{\sqrt {1 + {x^2}} }}$
C
$1 + {x^2}$
D
None of these

✓

Answer

Correct option: B.

$\frac{1}{{\sqrt {1 + {x^2}} }}$

b
(b) Let $\theta = {\tan ^{ - 1}}x\,\, \Rightarrow \,\,x = \tan \theta $

$\therefore \,\,\cos \theta = \frac{1}{{\sqrt {1 + {{\tan }^2}\theta } }} = \frac{1}{{\sqrt {1 + {x^2}} }}$

Hence $\cos \theta = \cos \,({\tan ^{ - 1}}x) = \frac{1}{{\sqrt {1 + {x^2}} }}$.

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