A) \[\frac{1}{2x}\]
B) \[{{x}^{2}}\]
C) \[\frac{1}{{{x}^{2}}}\]
D) \[\frac{2}{x}\]
Correct Answer: C
Solution :
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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