<s>[INST] If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy} {ab}\cos(\theta)$ If $\cos^{-1}(\frac{x}{a})+\cos^{-1}(\frac{y}{b})=\theta$, prove 
that $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos(\theta)=\sin^2(\theta)$
My trial:Let:
$\cos^{-1}(\frac{x}{a})=\alpha$,  and $\cos^{-1}(\frac{x}{a})=\beta$
$\sin(\theta)= \sin(\alpha) \cos(\beta)+\cos(\alpha)\sin(\beta)$
$\sin (\theta)=\sqrt{1-\frac{y^2}{b^2}}\bigl(\frac{x}{a}\big)+\sqrt{1-\frac{x^2}{a^2}}\big(\frac{y}{b}\bigr)$
$\sin^2(\theta)=\bigl(1-\frac{y^2}{b^2}\bigr)\frac{x^2}{a^2}+\bigl(1-\frac{x^2}{a^2}\bigr)\frac{y^2}{b^2}+\frac{2xy}{ab}\sqrt{(1-\frac{y^2}{b^2})((1-\frac{x^2}{a^2})}$
$\sin^2(\theta)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{2xy}{ab}\sqrt{(1-\frac{y^2}{b^2})((1-\frac{x^2}{a^2})}-\frac{2x^2y^2}{a^2b^2}$
I could not get the final solution, any idea, maybe there is a better algorithm?
 [/INST] Hint
Using $\cos(A+B)$ formula
$$\cos\theta-\dfrac{xy}{ab}=-\sqrt{\left(1-\dfrac{x^2}{a^2}\right)(\cdots)}$$
Take square in both sides
Alternatively 
Let $\cos^{-1}\dfrac xa=A,\cos A=?$ etc.
$$\cos\theta=\cos(A+B)$$
Rearrange and square both sides
$$(\cos\theta-\cos A\cos B)^2=(-\sin A\sin  B)^2=(1-\cos^2A)(1-\cos^2B)$$
Replace the values of $\cos A,\cos B$ to eliminate foreign elements $A,B$
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