MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\sin \left(\cot ^{-1} x\right)\) is
- A \(\frac{1}{\sqrt{1+x^2}}\)
- B \(\sqrt{1+x^2}\)
- C \(\frac{1}{x \sqrt{1+x^2}}\)
- D \(x \sqrt{1+x^2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sqrt{1+x^2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll} & \sin \left(\cot ^{-1} x\right) \\ & \text { Let } \cot ^{-1} x=\mathrm{t} \\ \therefore \quad & x=\cot \mathrm{t} \\ \therefore \quad & 1+\cot ^2 \mathrm{t}=1+x^2 \\ \therefore \quad & \operatorname{cosec}^2 \mathrm{t}=1+x^2 \\ \therefore \quad & \operatorname{cosec} \mathrm{t}=\sqrt{1+x^2} \\ \therefore \quad & \sin \mathrm{t}=\frac{1}{\sqrt{1+x^2}} \\ \therefore \quad & \mathrm{t}=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \\ \therefore \quad & \sin \left(\cot ^{-1} x\right)=\sin \left(\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right) \\ & =\frac{1}{\sqrt{1+x^2}}\end{array}\)
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