JEE Main 2025 Apr 2 Shift 2, Mathematics Q23

Let $\varphi ={\cos }^{-1}\frac{x}{2}$, so $\cos \varphi =\frac{x}{2}$ and $\sin \varphi =\frac{\sqrt{4-x^2}}{2}$.

Apply the cosine sum formula:

$y=\cos \frac{\pi }{3}\cos \varphi -\sin \frac{\pi }{3}\sin \varphi =\frac{1}{2}\cdot \frac{x}{2}-\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{4-x^2}}{2}=\frac{x-\sqrt{3}\sqrt{4-x^2}}{4}$.

Rearrange: $4y-x=-\sqrt{3}\sqrt{4-x^2}$. Squaring:

$(4y-x{)}^2=3(4-x^2)\Rightarrow 16y^2-8xy+x^2=12-3x^2$.

$\Rightarrow 4x^2+16y^2-8xy=12\Rightarrow x^2+4y^2-2xy=3$.

Now $(x-y{)}^2+3y^2=x^2-2xy+y^2+3y^2=x^2+4y^2-2xy=3$.