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