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Answer: $\sin ^{2} \alpha$

Hint: Let $\cos \alpha$ get into RHS, so that variable can free

Given:

If,$\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$

$\frac{x^{2}}{a^{2}}-\frac{2 x y}{a b} \cos \alpha+\frac{y^{2}}{b^{2}}=?$

Solution:

We know that,

\begin{aligned} &\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right. \\ &\therefore \cos ^{-1}\left(\frac{x}{a}\right)+\cos ^{-1}\left(\frac{y}{b}\right)=\alpha \end{aligned}

\begin{aligned} &\cos ^{-1}\left(\frac{x}{a} \frac{y}{b}-\sqrt{1-\frac{x^{2}}{a^{2}}} \sqrt{1-\frac{y^{2}}{b^{2}}}\right)=\alpha \\ &\frac{x y}{a b}-\sqrt{1-\frac{x^{2}}{a^{2}}} \sqrt{1-\frac{y^{2}}{b^{2}}}=\cos \alpha \\ &\sqrt{1-\frac{x^{2}}{a^{2}}} \sqrt{1-\frac{y^{2}}{b^{2}}}=\frac{x y}{a b}-\cos \alpha \end{aligned}

Squaring on both the sides,

$\left(1-\frac{x^{2}}{a^{2}}\right)\left(1-\frac{y^{2}}{b^{2}}\right)=\frac{x^{2}}{y^{2}} \frac{y^{2}}{b^{2}}+\cos ^{2} \alpha-\frac{2 x y}{a b} \cos \alpha$        Mention a square instead of y square in denominator

\begin{aligned} &1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}+\frac{x^{2}}{a^{2}} \frac{y^{2}}{b^{2}}=\frac{x^{2}}{a^{2}} \frac{y^{2}}{b^{2}} \\ &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{2 x y}{a b} \cos \alpha=1-\cos ^{2} \alpha \end{aligned}

$=\sin ^{2} \alpha$                  MEntion how the elimination of last two term has been done

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