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Question: Answered & Verified by Expert
The value of $\sin \left(\cot ^{-1} x\right)$ is
MathematicsInverse Trigonometric FunctionsMHT CETMHT CET 2023 (12 May Shift 1)
Options:
  • A $\frac{1}{\sqrt{1+x^2}}$
  • B $\sqrt{1+x^2}$
  • C $\frac{1}{x \sqrt{1+x^2}}$
  • D $x \sqrt{1+x^2}$
Solution:
2981 Upvotes Verified Answer
The correct answer is: $\frac{1}{\sqrt{1+x^2}}$
$\begin{array}{ll} & \sin \left(\cot ^{-1} x\right) \\ & \text { Let } \cot ^{-1} x=\mathrm{t} \\ \therefore \quad & x=\cot \mathrm{t} \\ \therefore \quad & 1+\cot ^2 \mathrm{t}=1+x^2 \\ \therefore \quad & \operatorname{cosec}^2 \mathrm{t}=1+x^2 \\ \therefore \quad & \operatorname{cosec} \mathrm{t}=\sqrt{1+x^2} \\ \therefore \quad & \sin \mathrm{t}=\frac{1}{\sqrt{1+x^2}} \\ \therefore \quad & \mathrm{t}=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \\ \therefore \quad & \sin \left(\cot ^{-1} x\right)=\sin \left(\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right) \\ & =\frac{1}{\sqrt{1+x^2}}\end{array}$

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