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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
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Verified Answer
The correct answer is:
$9\sin ^2 \alpha$
If $\cos ^{-1} \frac{x}{a}-\cos ^{-1} \frac{y}{b}=\theta$, then $\frac{x^2}{\mathrm{a}^2}-\frac{2 x y}{\mathrm{ab}} \cos \theta+\frac{y^2}{\mathrm{~b}^2}=\sin ^2 \theta$ Given, $\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha$ Here, $\mathrm{a}=1, \mathrm{~b}=3$
$\begin{aligned}
\therefore \quad & \frac{x^2}{1^2}-\frac{2 x y}{(1)(3)} \cos \alpha+\frac{y^2}{3^2}=\sin ^2 \alpha \\
& \Rightarrow x^2-\frac{2 x y}{3} \cos \alpha+\frac{y^2}{9}=\sin ^2 \alpha \\
& \Rightarrow 9 x^2-6 x y \cos \alpha+y^2=9 \sin ^2 \alpha
\end{aligned}$
$\begin{aligned}
\therefore \quad & \frac{x^2}{1^2}-\frac{2 x y}{(1)(3)} \cos \alpha+\frac{y^2}{3^2}=\sin ^2 \alpha \\
& \Rightarrow x^2-\frac{2 x y}{3} \cos \alpha+\frac{y^2}{9}=\sin ^2 \alpha \\
& \Rightarrow 9 x^2-6 x y \cos \alpha+y^2=9 \sin ^2 \alpha
\end{aligned}$
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