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CBSE Class 12 Math 2008 Solved Paper

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Question : 15 of 29
Marks: +1, -0
Differentiate the following with respect of x:
y = tan⁡−1(1+x−1−x1+x+1−x)\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)
Solution:  
Let x = cos 2θ ⇒ θ = 12cos⁡−1\frac{1}{2}\cos^{-1} x ... 1
∴ 1+x\sqrt{1+x} = 1+cos⁡2θ\sqrt{1+\cos 2\theta} = 1+2cos⁡2θ−1\sqrt{1+2\cos^2\theta-1} = 2\sqrt{2} cos θ
1−x\sqrt{1-x} = 1−cos⁡2θ\sqrt{1-\cos 2\theta} = 1−1−2sin⁡2θ\sqrt{1-1-2\sin^2\theta} = 2\sqrt{2} sin θ
Let y = tan⁡−1∣1+x−1−x1+x+1−x∣\tan^{-1}\left|\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right|
= tan⁡−1∣2cos⁡θ−2sin⁡θ2cos⁡θ+2sin⁡θ∣\tan^{-1}\left|\frac{\sqrt{2}\cos\theta-\sqrt{2}\sin\theta}{\sqrt{2}\cos\theta+\sqrt{2}\sin\theta}\right|
= tan⁡−1∣1−tan⁡θ1+tan⁡θ∣\tan^{-1}\left|\frac{1-\tan\theta}{1+\tan\theta}\right|
= tan⁡−1{tan⁡(π/4−θ)}\tan^{-1}\{\tan(\pi/4-\theta)\}
= π4\frac{\pi}{4} - θ = π4\frac{\pi}{4} - 12cos⁡−1\frac{1}{2}\cos^{-1} x From 1
∴ dydx\frac{dy}{dx} = −12(−11−x2)-\frac{1}{2}\left(-\frac{1}{\sqrt{1-x^2}}\right) = 121−x2\frac{1}{2\sqrt{1-x^2}}
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