JEE Main 2019 (Online) 10th April Evening ShiftQuestion 1
If $\cos^{-1}x-\cos^{-1}\frac y2=\alpha$, where $-1\le x\le1$, $-2\le y\le2$, $x\le\frac y2$, then for all $x,y$, $4x^{2}-4xy\cos\alpha+y^{2}$ is equal to :
Inverse Trigonometric Functions
Solution
With $\cos^{-1}x=A$, $\cos^{-1}\frac y2=B$: $\cos(A-B)=\cos\alpha$ gives $\frac{xy}2+\sqrt{1-x^2}\sqrt{1-\frac{y^2}4}=\cos\alpha$; isolating the root and squaring: $4x^2-4xy\cos\alpha+y^2=4\sin^2\alpha$.