Verified Solution • 2025–26 Session moderate

QJEEM24Ma1518: JEE Main 2024 Mathematics - Class 12 JEEM24Ma1518 Solution: Given that the inverse trigonometric function assumes principal …

Given that the inverse trigonometric function assumes principal values only. Let x,yx, y 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

Multiple Choice Options

  • A 0
  • B −- 1
  • C 12\frac{1}{2}
  • D −12\frac{-1}{2}

Step-by-Step Verified Solution

amp;cos⁡−1x−π2+cos⁡−1y=αamp;cos⁡−1x+cos⁡−1y=π2+αamp;∵α∈[−π2,π]amp; then π2+α∈(0,3π2)amp;cos⁡−1(xy−1−x21−y2)=π2+αamp;xy−1−x21−y2=−sin⁡αamp;xy+sin⁡α=1−x21−y2amp;x2y2+sin⁡2α+2xysin⁡α=1−x2−y2+x2y2amp;x2+y2+2xysin⁡α⏟E=cos⁡2α\begin{aligned}& \cos ^{-1} x-\frac{\pi}{2}+\cos ^{-1} y=\alpha \\& \cos ^{-1} x+\cos ^{-1} y=\frac{\pi}{2}+\alpha \\& \because \quad \alpha \in\left[\frac{-\pi}{2}, \pi\right] \\& \text { then } \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=\sqrt{1-x^2} \sqrt{1-y^2} \\& x^2 y^2+\sin ^2 \alpha+2 x y \sin \alpha=1-x^2-y^2+x^2 y^2 \\& \underbrace{x^2+y^2+2 x y \sin \alpha}_E=\cos ^2 \alpha\end{aligned}

Now, minimum value of EE is 0.

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