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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