MCQ
$\tan ({\cos ^{ - 1}}x)$ is equal to
  • $\frac{{\sqrt {1 - {x^2}} }}{x}$
  • B
    $\frac{x}{{1 + {x^2}}}$
  • C
    $\frac{{\sqrt {1 + {x^2}} }}{x}$
  • D
    $\sqrt {1 - {x^2}} $

Answer

Correct option: A.
$\frac{{\sqrt {1 - {x^2}} }}{x}$
a
(a) Let ${\cos ^{ - 1}}x = \theta .$ Then $x = \cos \theta $

$ \Rightarrow \,\,\tan \theta = \sqrt {{{\sec }^2}\theta - 1} $

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

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

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