Mathematics · Inverse Trigonometric Functions

JEE Main 2025 — 2 April, Evening Shift — Question 44

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

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

y=cos⁡(π3+cos⁡−1x2)y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right)

=cos⁡(π3)cos⁡(cos⁡−1(x2))−sin⁡(π3)sin⁡(cos⁡−1(x2))=12⋅x2−32⋅1−x24⇒4y=x−34−x2⇒(4y−x)2=3(4−x2)⇒16y2+x2−8xy=12−3x2x2+4y2−2xy=3(x−y)2+3y2=3\begin{aligned} & =\cos \left(\frac{\pi}{3}\right) \cos \left(\cos ^{-1}\left(\frac{x}{2}\right)\right)-\sin \left(\frac{\pi}{3}\right) \sin \left(\cos ^{-1}\left(\frac{x}{2}\right)\right) \\& =\frac{1}{2} \cdot \frac{x}{2}-\frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^{2}}{4}} \\& \Rightarrow 4 y=x-\sqrt{3} \sqrt{4-x^{2}} \\& \Rightarrow(4 y-x)^{2}=3\left(4-x^{2}\right) \\& \Rightarrow 16 y^{2}+x^{2}-8 x y=12-3 x^{2} \\& x^{2}+4 y^{2}-2 x y=3 \\& (x-y)^{2}+3 y^{2}=3 \end{aligned}

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Exam
JEE Main 2025
Subject
Mathematics
Chapter
Inverse Trigonometric Functions
Topic
Properties related to Inverse Trigonometric Functions