MHT CET · Maths · Inverse Trigonometric Functions
If \(y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\sec ^{-1}\left(\frac{1+x^2}{1-x^2}\right)\) then the value of \(\frac{\mathrm{dy}}{\mathrm{d} x}\) at \(x=\sqrt{3}\) is
- A \(1\)
- B \(\frac{1}{2}\)
- C \(0\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(C) \(0\)
Step-by-step Solution
Detailed explanation
Let \(x=\tan\theta\). Since \(x=\sqrt{3}>1\), then \(\theta \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)\). \(\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin 2\theta)\). For \(2\theta \in \left(\frac{\pi}{2}, \pi\right)\), this is \(\pi - 2\theta = \pi - 2\tan^{-1}x\).
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