JEE Main 2025 (Online) 2nd April Evening ShiftQuestion 1
If $y=\cos\left(\frac{\pi}{3}+\cos^{-1}\frac{x}{2}\right)$, then $(x-y)^{2}+3y^{2}$ is equal to $\_\_\_\_\_\_$ .
Inverse Trigonometric Functions
Answer$3$
Solution
Let $s=\sqrt{4-x^2}$, so $\sin\cos^{-1}\frac x2=\frac s2$. $y=\frac12\cdot\frac x2-\frac{\sqrt3}2\cdot\frac s2=\frac{x-\sqrt3s}4$. $x-y=\frac{3x+\sqrt3s}4$. $(x-y)^2+3y^2=\frac{(3x+\sqrt3s)^2+3(x-\sqrt3s)^2}{16}=\frac{9x^2+3s^2+3x^2+9s^2}{16}=\frac{12(x^2+s^2)}{16}=\frac{12\cdot4}{16}=3$.