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← JEE Main 2017 · 8 Apr · Shift 1
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JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 89

The value of tan⁡−1[1+x2+1−x21+x2−1−x2]\tan^{-1}\left[\dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right], ∣x∣<12|x|<\dfrac12, x≠0x\ne0, is equal to :

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Answer and worked solution

Official answer: A

Put x2=cos⁡2θx^2=\cos2\theta. Then the fraction is cos⁡θ+sin⁡θcos⁡θ−sin⁡θ=tan⁡(π4+θ)\dfrac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=\tan\left(\frac\pi4+\theta\right), so the value is
π4+θ=π4+12cos⁡−1x2.\frac\pi4+\theta=\frac\pi4+\frac12\cos^{-1}x^2.

Topics:Inverse trigonometric functions
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