Question

$\frac{dy}{dx}$ for $y={\tan }^{-1}\left\{\sqrt{\frac{1+\cos x}{1-\cos x}}\right\}$, where $0<x<\pi$, is?

• A. $\frac{-1}{2}$
• B. $0$
• C. $1$
• D. $-1$

Answer 1

Answer: (A)

$\frac{-1}{2}$

$Tofind\frac{d}{dx}\left({\tan }^{-1}\sqrt{\frac{1+\cos x}{1-\cos x}}\right)ByApplyingchainruleletustakeu=\sqrt{\frac{1+\cos x}{1-\cos x}}Therefore\frac{d}{du}\left({\tan }^{-1}u\right)\cdot \frac{d}{dx}\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)Step1:\frac{d}{du}{\tan }^{-1}u=\frac{1}{u^2+1}Step2:\frac{d}{dx}\sqrt{\frac{1+\cos x}{1-\cos x}}=-\frac{\sin x}{\sqrt{1+\cos x}\cdot ({1-\cos x)}^{3/2}}CombiningStep1andStep2=\frac{1}{u^2+1}\cdot (-\frac{\sin x}{\sqrt{1+\cos x}\cdot ({1-\cos x)}^{3/2}})substitutingthevalueof{}^{\prime }{u}^{\prime }inthegivenequation=\frac{1}{{\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)}^2+1}\cdot (-\frac{\sin x}{\sqrt{1+\cos x}\cdot ({1-\cos x)}^{3/2}})=-\frac{sinx}{2\sqrt{1+cosx}(1-{cosx)}^{1/2}}Since0<x<\pi Letustakethevalueas\frac{\pi }{2}Sothevaluewouldbe=-\frac{\sin \frac{\pi }{2}}{2\sqrt{1+\cos \frac{\pi }{2}}{(1-\cos \frac{\pi }{2})}^{\frac{1}{2}}}=-\frac{1}{2\sqrt{1+0}(1-0{)}^{1/2}}=-\frac{1}{2}Therefore\frac{d}{dx}\left({\tan }^{-1}\sqrt{\frac{1+\cos x}{1-\cos x}}\right)=-\frac{1}{2}at0<x<\pi$

Hence, $\frac{-1}{2}$ is the correct answer.