MCQ
$1 + {\cot ^2}({\sin ^{ - 1}}x) = $
  • A
    $\frac{1}{{2x}}$
  • B
    ${x^2}$
  • $\frac{1}{{{x^2}}}$
  • D
    $\frac{2}{x}$

Answer

Correct option: C.
$\frac{1}{{{x^2}}}$
c
(c) Let ${\sin ^{ - 1}}x = \theta \,\, \Rightarrow \,\,\sin \theta = x$

Now $1 + {\cot ^2}\theta = \cos e{c^2}\theta = \frac{1}{{{x^2}}}$

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

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