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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Inverse Trigonometric Functions

Given that the inverse trigonometric function assumes principal...

Given that the inverse trigonometric function assumes principal values only. Let xx, yy be any two real numbers in [−1,1][-1,1] such that cos⁡−1x−sin⁡−1y=α,−π2≤α≤π\cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi.
Then, the minimum value of x2+y2+2xysin⁡αx^2+y^2+2 x y \sin \alpha is

Options

  1. A0
  2. B-1
  3. C12\frac{1}{2}
  4. D−12-\frac{1}{2}

Correct answer

A. 0

Step-by-step solution

cos⁡−1x−(π2−cos⁡−1y)=αcos⁡−1x+cos⁡−1y=π2+αα∈[−π2,π],π2+α∈[0,3π2]cos⁡−1(xy−1−x21−y2)=π2+αxy−1−x21−y2=−sin⁡α(xy+sin⁡α)=(1−x2)(1−y2)x2y2+2xysin⁡a+sin⁡2a=1−x2−y2+x2y2x2+y2+2xysin⁡α=1−sin⁡2αx2+y2+2xysin⁡α=cos⁡2α\begin{aligned} & \cos ^{-1} x-\left(\frac{\pi}{2}-\cos ^{-1} y\right)=\alpha \\ & \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\ & \alpha \in\left[-\frac{\pi}{2}, \pi\right], \frac{\pi}{2}+\alpha \in\left[0, \frac{3 \pi}{2}\right] \\ & \cos ^{-1}\left(x y-\sqrt{1-x^2} \sqrt{1-y^2}\right)=\frac{\pi}{2}+\alpha \\ & x y-\sqrt{1-x^2} \sqrt{1-y^2}=-\sin \alpha \\ & (x y+\sin \alpha)=\left(1-x^2\right)\left(1-y^2\right) \\ & x^2 y^2+2 x y \sin a+\sin ^2 a=1-x^2-y^2+x^2 y^2 \\ & x^2+y^2+2 x y \sin \alpha=1-\sin ^2 \alpha \\ & x^2+y^2+2 x y \sin \alpha=\cos ^2 \alpha \end{aligned}
Min. value of cos⁡2α=0\cos ^2 \alpha=0 At α=π2\alpha=\frac{\pi}{2}
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