MHT CET · Maths · Differentiation
If \(y=\log \tan \left(\frac{x}{2}\right)+\sin ^1(\cos x)\), then \(\frac{d y}{d x}=\)
- A \(\operatorname{cosec} x\)
- B \(\sin x+1\)
- C x
- D \(\operatorname{cosec} x-1\)
Answer & Solution
Correct Answer
(D) \(\operatorname{cosec} x-1\)
Step-by-step Solution
Detailed explanation
\(y=\log \tan \left(\frac{x}{2}\right)+\sin ^{-1}(\cos x) \)
\( =\log \tan \left(\frac{x}{2}\right)+\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-x\right)\right]=\log \tan \left(\frac{x}{2}\right)+\) \(\frac{\pi}{2}-x \)
\( \therefore \frac{d y}{d x}=\frac{1}{\tan \left(\frac{x}{2}\right)} \times \sec ^2 \times\left(\frac{x}{2}\right) \times\left(\frac{1}{2}\right)+0-1 \)
\( =\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \times \frac{1}{\cos ^2 \frac{x}{2}} \times \frac{1}{2}-1 \)
\( =\frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}-1=\frac{1}{\sin x}-1=\operatorname{cosec} x-1\)
\( =\log \tan \left(\frac{x}{2}\right)+\sin ^{-1}\left[\sin \left(\frac{\pi}{2}-x\right)\right]=\log \tan \left(\frac{x}{2}\right)+\) \(\frac{\pi}{2}-x \)
\( \therefore \frac{d y}{d x}=\frac{1}{\tan \left(\frac{x}{2}\right)} \times \sec ^2 \times\left(\frac{x}{2}\right) \times\left(\frac{1}{2}\right)+0-1 \)
\( =\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}} \times \frac{1}{\cos ^2 \frac{x}{2}} \times \frac{1}{2}-1 \)
\( =\frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}-1=\frac{1}{\sin x}-1=\operatorname{cosec} x-1\)
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