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  • <img alt="\frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1-\cos(x)}{1+\cos (x)}}\right)" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bd%7D%7Bd%20x%7D%5Cleft%28%5Ctan%20%5E%7B-1%7D%20%5

\frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1-\cos(x)}{1+\cos (x)}}\right) is equal to (Given x \in (0,\pi))

Option: 1

-\frac{1}{4}


Option: 2

{\frac{1}{2}}


Option: 3

{-\frac{1}{2}}


Option: 4

{\frac{1}{4}}


Answers (1)

best_answer

In many questions involving differentiation of complex inverse trigonometric functions, we need to first simplify the given expression using techniques learnt in the concept 'Simplification using Substitution' in the chapter 'Inverse Trigonometric Functions'.

\\\text{Let }y=\tan ^{-1} \sqrt{\frac{1-\cos (x)}{1+\cos (x)}}\\ =\tan ^{-1} \sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}\\ =\tan ^{-1} |\tan \frac{x}{2}|\\ As\,\,x\in(0,\pi), so\,\,\frac{x}{2} \in (0,\frac{\pi}{2}), and \,\,tan(\frac{x}{2})\,\,is\,\,positive \\\\ =\tan ^{-1} \tan \left(\frac{x}{2}\right)=\frac{x}{2}\\\therefore \frac{dy}{dx}=\frac{1}{2}

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