Question

If ${\cos }^{-1}\frac{x}{2}+{\cos }^{-1}\frac{y}{3}=\mathrm{\alpha },$ then prove that 9x² − 12xy cos α + 4y² = 36 sin² α.

Answer 1

We know
${\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}\left[xy-\sqrt{1-x^2}\sqrt{1-y^2}\right]$
Now,

$$\begin{gathered} {\cos }^{-1}\frac{x}{2}+{\cos }^{-1}\frac{y}{3}=\alpha \\ \Rightarrow {\cos }^{-1}\left[\frac{x}{2}\frac{y}{3}-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{3}}\right]=\alpha \\ \Rightarrow \frac{x}{2}\frac{y}{3}-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{3}}=\cos \alpha \\ \Rightarrow xy-\sqrt{4-x^2}\sqrt{9-y^2}=6\cos \alpha \\ \Rightarrow \sqrt{4-x^2}\sqrt{9-y^2}=xy-6\cos \alpha \\ \Rightarrow \left(4-x^2\right)\left(9-y^2\right)=x^2{y}^2+36{\cos }^2\alpha -12xy\cos \alpha \left[\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right] \\ \Rightarrow 36-4y^2-9x^2+x^2{y}^2=x^2{y}^2+36{\cos }^2\alpha -12xy\cos \alpha \\ \Rightarrow 36-4y^2-9x^2=36{\cos }^2\alpha -12xy\cos \alpha \\ \Rightarrow 9x^2-12xy\cos \alpha +4y^2=36-36{\cos }^2\alpha \\ \Rightarrow 9x^2-12xy\cos \alpha +4y^2=36{\sin }^2\alpha \end{gathered}$$