MHT CET · Maths · Inverse Trigonometric Functions
If \(y=\tan ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right]\), then \(\frac{d y}{d x}=\)
- A \(\frac{-1}{3}\)
- B \(\frac{-1}{4}\)
- C \(\frac{1}{3}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{-1}{4}\)
Step-by-step Solution
Detailed explanation
Given \(y=\tan ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right]\)
\(
=\tan ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{x}{4}}{2 \sin ^{2} \frac{x}{4}}}=\tan ^{-1}\left(\cot \frac{x}{4}\right)=\tan ^{-1}\) \(\left[\tan \left(\frac{\pi}{2}-\frac{x}{4}\right)\right]
\)
\(\therefore y=-\frac{x}{4} \Rightarrow \frac{d y}{d x}=\frac{-1}{4}\)
\(
=\tan ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{x}{4}}{2 \sin ^{2} \frac{x}{4}}}=\tan ^{-1}\left(\cot \frac{x}{4}\right)=\tan ^{-1}\) \(\left[\tan \left(\frac{\pi}{2}-\frac{x}{4}\right)\right]
\)
\(\therefore y=-\frac{x}{4} \Rightarrow \frac{d y}{d x}=\frac{-1}{4}\)
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