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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Inverse Trigonometric Functions

If y= ((π)/(3)+ ^ -1 (x)/(2)) , then (x-y)^2+3 y^2 is equal to...

If y=cos⁡(π3+cos⁡−1x2)y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right), then (x−y)2+3y2(x-y)^2+3 y^2 is equal to _____.

Correct answer

0

Step-by-step solution

y=cos⁡(cos⁡−112+cos⁡−1x2)y=12×x2−1−141−x244y=x−34−x23(4−x2)=x2+16y2−8xy12−3x2=x2+16y2−8xy4x2+16y2−8xy=12x2+4y2−2xy=3x2+y2−2xy−3y2=3(x−y)2+3y2=3\begin{aligned} & y=\cos \left(\cos ^{-1} \frac{1}{2}+\cos ^{-1} \frac{x}{2}\right) \\ & y=\frac{1}{2} \times \frac{x}{2}-\sqrt{1-\frac{1}{4}} \sqrt{1-\frac{x^2}{4}} \\ & 4 y=x-\sqrt{3} \sqrt{4-x^2} \\ & 3\left(4-x^2\right)=x^2+16 y^2-8 x y \\ & 12-3 x^2=x^2+16 y^2-8 x y \\ & 4 x^2+16 y^2-8 x y=12 \\ & x^2+4 y^2-2 x y=3 \\ & x^2+y^2-2 x y-3 y^2=3 \\ & (x-y)^2+3 y^2=3\end{aligned}
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