If ${\cos }^{-1}\frac{p}{a}+{\cos }^{-1}\frac{q}{b}=\alpha$ then prove that $\frac{p^2}{a^2}-2\frac{pq}{ab}\cos \alpha +\frac{q^2}{b^2}={\sin }^2\alpha$

Let ${\cos }^{-1}\frac{p}{a}=A\Rightarrow \cos A=\frac{p}{a}$

${\cos }^{-1}\frac{q}{b}=B\Rightarrow \cos B=1\frac{q}{b}$

Now, ${\cos }^{-1}\frac{p}{a}+{\cos }^{-1}\frac{q}{b}=\alpha \Rightarrow A+B=\alpha$

$\cos \left(\alpha \right)=\cos (A+B)$

$\cos \alpha =\cos A\cos B-\sin A\sin B$

$\cos \alpha =\frac{p}{a}\cdot \frac{q}{b}-\sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}}$

$\Rightarrow \sqrt{1-\frac{p^2}{a^2}}\sqrt{1-\frac{q^2}{b^2}}=\frac{pq}{ab}-\cos \alpha$

S.O.B.S. 

$\Rightarrow \left(1-\frac{p^2}{a^2}\right)\left(1-\frac{q^2}{b^2}\right)={\left(\frac{pq}{ab}-\cos \alpha \right)}^2$$\Rightarrow 1-\frac{p^2}{a^2}-\frac{q^2}{b^2}+\frac{p^2{q}^2}{a^2{b}^2}=\frac{p^2{q}^2}{a^2{b}^2}-\frac{2pq}{ab}\cos \alpha +{\cos }^2\alpha$

$\therefore$$\frac{p^2}{a^2}+\frac{q^2}{b^2}-\frac{2pq}{ab}\cos \alpha =1-{\cos }^2\alpha ={\sin }^2\alpha$