MHT CET · Maths · Inverse Trigonometric Functions
The value of \(x\), for which \(\sin \left(\cot ^{-1}(x)\right)=\cos \left(\tan ^{-1}(1+x)\right)\), is
- A \(0\)
- B \(1\)
- C \(-\frac{1}{2}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Note that \(\cot ^{-1} x=\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\) and
\(\tan ^{-1}(1+x)=\cos ^{-1}\left(\frac{1}{\sqrt{1+(1+x)^2}}\right) \)
\( \therefore \quad \sin \left(\cot ^{-1}(x)\right)=\cos \left(\tan ^{-1}(1+x)\right) \)
\( \Rightarrow \sin \left(\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right)=\) \(\cos \left(\cos ^{-1}\left(\frac{1}{\sqrt{1+(1+x)^2}}\right)\right) \)
\( \Rightarrow \frac{1}{\sqrt{1+x^2}}=\frac{1}{\sqrt{1+(1+x)^2}} \)
\( \Rightarrow 1+(1+x)^2=1+x^2 \)
\( \Rightarrow x=\frac{-1}{2}\)
\(\tan ^{-1}(1+x)=\cos ^{-1}\left(\frac{1}{\sqrt{1+(1+x)^2}}\right) \)
\( \therefore \quad \sin \left(\cot ^{-1}(x)\right)=\cos \left(\tan ^{-1}(1+x)\right) \)
\( \Rightarrow \sin \left(\sin ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right)=\) \(\cos \left(\cos ^{-1}\left(\frac{1}{\sqrt{1+(1+x)^2}}\right)\right) \)
\( \Rightarrow \frac{1}{\sqrt{1+x^2}}=\frac{1}{\sqrt{1+(1+x)^2}} \)
\( \Rightarrow 1+(1+x)^2=1+x^2 \)
\( \Rightarrow x=\frac{-1}{2}\)
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