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CBSE Class 12 Math 2022 Term I Solved Paper

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Question : 31 of 50
Marks: +1, -0
The simplest form of 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] is
Solution:     👈: Video Solution
Explanation: We have,
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)
Put x=cos⁡2θx=\cos 2\theta, so that θ=12cos⁡−1x\theta=\frac{1}{2} \cos^{-1} x
tan⁡−1(1+cos⁡2θ−1−cos⁡2θ1+cos⁡2θ+1−cos⁡2θ)\tan^{-1} \left( \frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right)
=tan⁡−1(2cos⁡2θ−2sin⁡2θ2cos⁡2θ+2sin⁡2θ)=\tan^{-1} \left( \frac{\sqrt{2\cos^2 \theta} - \sqrt{2\sin^2 \theta}}{\sqrt{2\cos^2 \theta} + \sqrt{2\sin^2 \theta}} \right)
=tan⁡−1(cos⁡θ−sin⁡θcos⁡θ+sin⁡θ)=\tan^{-1} \left( \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta} \right)
=tan⁡−1(1−tan⁡θ1+tan⁡θ)=\tan^{-1} \left( \frac{1 - \tan \theta}{1 + \tan \theta} \right)
=tan⁡−1(1)−tan⁡−1(tan⁡θ)=\tan^{-1}(1) - \tan^{-1}(\tan \theta)
=tan⁡−1(tan⁡4)−θ=\tan^{-1} (\tan^4) - \theta
=π4−12cos⁡−1x=\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x
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