Question

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

Answer 1

${\displaystyle {\cos }^{-1}\left(\frac{x}{a}\right)+{\cos }^{-1}\left(\frac{y}{b}\right)=\alpha }$ 

${\displaystyle {\cos }^{-1}\left(\frac{x}{a}\right)=\alpha -{\cos }^{-1}\left(\frac{y}{b}\right)}$

${\displaystyle \Rightarrow \cos \left\{{\cos }^{-1}\left(\frac{x}{a}\right)\right\}=\cos \{\alpha -{\cos }^{-1}\left(\frac{y}{b}\right)\}}$

${\displaystyle \Rightarrow \frac{x}{a}=\cos \alpha \cos \left\{{\cos }^{-1}\left(\frac{y}{b}\right)\right\}+\sin \alpha \sin \left\{{\cos }^{-1}\left(\frac{y}{b}\right)\right\}}$

${\displaystyle \Rightarrow \frac{x}{a}=\frac{y}{b}\cos \alpha +\sin \alpha \sqrt{1-{\left(\frac{y}{b}\right)}^2}}$

${\displaystyle \Rightarrow \frac{x}{a}-\frac{y}{b}\cos \alpha =\sin \alpha \sqrt{1-{\left(\frac{y}{b}\right)}^2}}$

Squaring both sides, we get

${\displaystyle {(\frac{x}{a}-\frac{y}{b}\cos \alpha )}^2={\left\{\sin \alpha \sqrt{1-{\left(\frac{y}{b}\right)}^2}\right\}}^2}$

${\displaystyle {\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2{\cos }^2\alpha -\frac{2xy}{ab}\cos \alpha ={\sin }^2\alpha -{\sin }^2\alpha {\left(\frac{y}{b}\right)}^2}$

${\displaystyle {\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2{\cos }^2\alpha +{\sin }^2\alpha {\left(\frac{y}{b}\right)}^2-\frac{2xy}{ab}\cos \alpha ={\sin }^2\alpha }$

${\displaystyle {\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2({\cos }^2\alpha +{\sin }^2\alpha )-\frac{2xy}{ab}\cos \alpha ={\sin }^2\alpha }$

${\displaystyle \Rightarrow {\left(\frac{x}{a}\right)}^2-\frac{2xy}{ab}\cos \alpha +{\left(\frac{y}{b}\right)}^2={\sin }^2\alpha }$