If $\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha$, where $-1 \leq x \leq 1$, $-3 \leq y \leq 3, x \leq…
If $\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha$, where $-1 \leq x \leq 1$, $-3 \leq y \leq 3, x \leq \frac{y}{3}$, then for all $x, y$ $9 x^2-6 x y \cos \alpha+y^2$ is equal to
$\sin ^2 \alpha$
$3\sin ^2 \alpha$
$9\sin ^2 \alpha$
$\frac{4}{9} \sin ^2 \alpha$
Solution
$\begin{aligned}
& \cos ^{-1} a-\cos ^{-1} b=\cos ^{-1}\left(a b+\sqrt{1-a^2} \cdot \sqrt{1-b^2}\right) \\
& \therefore \quad \cos ^{-1} x-\cos ^{-1} \frac{y}{3} \\
& =\cos ^{-1}\left(\frac{x y}{3}+\sqrt{1-x^2} \cdot \sqrt{1-\frac{y^2}{9}}\right)=\alpha \\
& \therefore \quad \frac{x y}{3}+\frac{\sqrt{1-x^2} \cdot \sqrt{9-y^2}}{3}=\cos \alpha \\
& \quad x y+\sqrt{1-x^2} \cdot \sqrt{9-y^2}=3 \cos \alpha \\
& x y-3 \cos \alpha=-\sqrt{1-x^2} \cdot \sqrt{9-y^2}
\end{aligned}$
squaring on both sides, we get
$\begin{aligned}
& x^2 y^2-6 x y \cos \alpha+9 \cos ^2 \alpha=\left(1-x^2\right)\left(9-y^2\right) \\
& x^2 y^2-6 x y \cos \alpha+9 \cos ^2 \alpha=9-y^2-9 x^2+x^2 y^2
\end{aligned}$
i.e., $9 x^2-6 x y \cos \alpha+y^2=9 \sin ^2 \alpha$