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CBSE Class 12 Math 2018 Solved Paper

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Question : 7 of 29
Marks: +1, -0
Differentiate tan⁡−1(1+cos⁡xsin⁡x)\tan^{-1}\left(\frac{1+\cos x}{\sin x}\right) with respect to x
Solution:  
Given y = tan⁡−1(1+cos⁡xsin⁡x)\tan^{-1}\left(\frac{1+\cos x}{\sin x}\right)
⇒ y = tan⁡−1\tan^{-1} (2cos⁡2x22sin⁡x2cos⁡x2)\left(\frac{2\cos^2 \frac{x}{2}}{2\sin \frac{x}{2} \cos \frac{x}{2}}\right) (Since 1 + cos x = 2 cos⁡2x2\cos^2 \frac{x}{2} and sin x = 2 sin x2\frac{x}{2} cos x2\frac{x}{2})
⇒ y = tan⁡−1\tan^{-1} (cot⁡x2)\left(\cot \frac{x}{2}\right)
⇒ y = tan⁡−1\tan^{-1} [tan⁡(π2−x2)]\left[ \tan\left(\frac{\pi}{2} - \frac{x}{2}\right) \right]
y = π2−x2\frac{\pi}{2} - \frac{x}{2} [Since tan⁡−1\tan^{-1} (tan x) = x]
Differentiating with respect to x,
⇒ dydx\frac{dy}{dx} = 0 - 12\frac{1}{2} = - 12\frac{1}{2}
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