CUET · MATHS · PYQ PAPER 2023
If \(y=\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\) then \(\frac{d y}{d x}=\)
- A \(-\frac{2}{1+x^2}\)
- B \(\frac{2}{1+x^2}\)
- C \(-\frac{1}{1+x^2}\)
- D \(\frac{1}{1+x^2}\)
Answer & Solution
Correct Answer
(A) \(-\frac{2}{1+x^2}\)
Step-by-step Solution
Detailed explanation
Let \(x=\tan\theta\). \(y=\sin^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)=\sin^{-1}(\cos(2\theta))\) \(y=\sin^{-1}\left(\sin\left(\frac{\pi}{2}-2\theta\right)\right)=\frac{\pi}{2}-2\theta\) \(y=\frac{\pi}{2}-2\tan^{-1}x\)…
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