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Question: \(\frac{d}{dx}\left\lbrack \sin^{2}{\cot^{- 1}\left\{ \sqrt{\frac{1 - x}{1 + x}} \right\}} \right\rb...

ddx[sin⁡2cot⁡−1{1−x1+x}]\frac{d}{dx}\left\lbrack \sin^{2}{\cot^{- 1}\left\{ \sqrt{\frac{1 - x}{1 + x}} \right\}} \right\rbrack equals

A

−1- 1

B

12\frac{1}{2}

C

−12- \frac{1}{2}

D

1

Answer

12\frac{1}{2}

Explanation

Solution

Let y=sin⁡2cot⁡−1{1−x1+x}y = \sin^{2}{\cot^{- 1}\left\{ \sqrt{\frac{1 - x}{1 + x}} \right\}}

Put x=cos⁡θ⇒θ=cos⁡−1xx = \cos\theta \Rightarrow \theta = \cos^{- 1}x

⇒ y=sin⁡2cot⁡−1{1−cos⁡θ1+cos⁡θ}=sin⁡2cot⁡−1(tan⁡θ2)y = \sin^{2}{\cot^{- 1}\left\{ \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \right\}} = \sin^{2}{\cot^{- 1}\left( \tan\frac{\theta}{2} \right)}

⇒y=sin⁡2(π2−θ2)y = \sin^{2}\left( \frac{\pi}{2} - \frac{\theta}{2} \right)= cos⁡2θ2\cos^{2}\frac{\theta}{2}= 12(1+cos⁡θ)\frac{1}{2}(1 + \cos\theta) =12(1+x)\frac{1}{2}(1 + x)

∴\therefore dydx=12\frac{dy}{dx} = \frac{1}{2}