MHT CET · Maths · Differentiation
If \(y=\cot ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)\), then \(\frac{d y}{d x}=\)
- A \(\frac{1}{2}\)
- B \(-1\)
- C \(\frac{1}{3}\)
- D 1
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(y =\cot ^{-1} \sqrt{\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}}=\cot ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right) \)
\( =\cot ^{-1}[\frac{1-\tan \frac{x}{2}}{\left.1+\tan \frac{x}{2}\right)}=\tan ^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)=\) \(\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \tan \frac{x}{2}}\right]\)
\( \therefore \frac{d y}{d x} =0+\frac{1}{2}=\frac{1}{2} \)
\( =\cot ^{-1}[\frac{1-\tan \frac{x}{2}}{\left.1+\tan \frac{x}{2}\right)}=\tan ^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)=\) \(\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \tan \frac{x}{2}}\right]\)
\( \therefore \frac{d y}{d x} =0+\frac{1}{2}=\frac{1}{2} \)
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