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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

If ^ -1 [ 1+x^2 - 1-x^2 1+x^2 + 1-x^2 ]=α , then the value of 2 α is

If tan⁡−1[1+x2−1−x21+x2+1−x2]=α\tan ^{-1}\left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right]=\alpha, then the value of sin⁡2α\sin 2 \alpha is

Options

  1. Ax3x^3
  2. Bx\sqrt{\mathrm{x}}
  3. Cx\mathrm{x}
  4. Dx2\mathrm{x}^2

Correct answer

D. x2\mathrm{x}^2

Step-by-step solution

WehaveWe have\tan \alpha= \left[\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right] Put x2=cos⁡θx^2=\cos \theta∴tan⁡α=2cos⁡2θ2−2sin⁡2θ22cos⁡2θ2+2sin⁡2θ2 and ∴tan⁡α=cos⁡θ2−sin⁡θ2cos⁡θ2−sin⁡θ2=1−tan⁡θ21+tan⁡θ2=tan⁡(π4−θ2) and ∴α=π4−θ2⇒2α=π2−θ and ∴sin⁡2α=sin⁡(π2−θ)=cos⁡θ=x2\begin{aligned}& \therefore \tan \alpha=\frac{\sqrt{2 \cos ^2 \frac{\theta}{2}-\sqrt{2 \sin ^2 \frac{\theta}{2}}}}{\sqrt{2 \cos ^2 \frac{\theta}{2}+\sqrt{2 \sin ^2 \frac{\theta}{2}}}}\\ \text{ and } \therefore \tan \alpha=\frac{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}=\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}=\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\\ \text{ and } \therefore \alpha=\frac{\pi}{4}-\frac{\theta}{2} \Rightarrow2 \alpha=\frac{\pi}{2}-\theta\\ \text{ and } \therefore \sin 2 \alpha=\sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta=x^2 \end{aligned}$
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