Question

If ${\cos }^{-1}\frac{x}{a}+{\cos }^{-1}\frac{y}{b}=\alpha$, prove that $\frac{x^2}{a^2}-\frac{2xy}{ab}\cos \alpha +\frac{y^2}{b^2}={\sin }^2\alpha$.

Answer 1

$\begin{array}{c}{\cos }^{-1}\frac{x}{a}+{\cos }^{-1}\frac{y}{b}=\alpha \left[\because {\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}\left(xy-\sqrt{\left(1-x^2\right)\left(1-y^2\right)}\right)\right]\\ {\cos }^{-1}\left[\frac{x}{a}\cdot \frac{y}{b}-\sqrt{\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)}\right]=\alpha \\ \cos \alpha =\frac{xy}{ab}-\sqrt{\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)}\\ \sqrt{\left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)}=\left(\frac{xy}{ab}-\cos \alpha \right)\\ Squaringonbothside\\ \left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)={\left(\frac{xy}{ab}\right)}^2-\frac{2xy}{ab}\cos \alpha +{\cos }^2\alpha \\ 1-\frac{y^2}{b^2}-\frac{x^2}{a^2}+{\left(\frac{xy}{ab}\right)}^2={\left(\frac{xy}{ab}\right)}^2-\frac{2xy}{ab}\cos \alpha +{\cos }^2\alpha \\ 1-{\cos }^2\alpha =\frac{y^2}{b^2}-\frac{2xy}{ab}\cos \alpha +\frac{x^2}{a^2}\\ {\sin }^2\alpha =\frac{y^2}{b^2}-\frac{2xy}{ab}\cos \alpha +\frac{x^2}{a^2}Proved\end{array}$