MCQ
$\sin ({\cot ^{ - 1}}x) =$
  • A
    $\sqrt {1 + {x^2}} $
  • B
    $x$
  • C
    ${(1 + {x^2})^{ - 3/2}}$
  • ${(1 + {x^2})^{ - 1/2}}$

Answer

Correct option: D.
${(1 + {x^2})^{ - 1/2}}$
d
(d) ${\cot ^{ - 1}}\left( {\frac{1}{x}} \right) = \theta \,\, \Rightarrow \,\,\cos \theta = \frac{1}{x}$

Now, $\cos ec \theta =\sqrt {1+ {\cot ^2 \theta}} = \sqrt {{x^2} +1} $.

$\therefore \,\,\,\sin \theta  = \frac{1}{{\cos ec\,\theta }} = \frac{1}{{\sqrt {1 + {x^2}} }}\,\, \Rightarrow \,\theta  = {\sin ^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}$

Hence $\sin \,({\cot ^{ - 1}}x)\, = \sin \,\left( {{{\sin }^{ - 1}}\frac{1}{{\sqrt {1 + {x^2}} }}} \right)$

$ = \frac{1}{{\sqrt {1 + {x^2}} }} = {(1 + {x^2})^{ - 1/2}}$

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