MHT CET · Maths · Inverse Trigonometric Functions
If \(\sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right)\) then considering positive square roots, \(x\) has the value \(\qquad\)
- A 0
- B \(\frac{9}{4}\)
- C \(\frac{1}{2}\)
- D \(-\frac{1}{2}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sin \left(\cot ^{-1}(x+1)\right)=\cos \left(\tan ^{-1} x\right) \\ & \sin \left[\sin ^{-1} \frac{1}{\sqrt{x^2+2 x+2}}\right]=\cos \left[\cos ^{-1} \frac{1}{\sqrt{1+x^2}}\right] \\ & \frac{1}{\sqrt{x^2+2 x+2}}=\frac{1}{\sqrt{1+x^2}} \\ & \Rightarrow \sqrt{1+x^2}=\sqrt{x^2+2 x+2} \\ & \Rightarrow 1+x^2=x^2+2 x+2 \\ & \Rightarrow 2 x+1=0 \\ & \Rightarrow x=\frac{-1}{2}\end{aligned}\)
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