Question

If ${\cos }^{-1}\frac{x}{2}+{\cos }^{-1}\frac{y}{3}=\mathrm{\theta },$ then 9x² − 12xy cos θ + 4y² is equal to
(a) 36
(b) −36 sin² θ
(c) 36 sin² θ
(d) 36 cos² θ

Answer 1

(c) 36 sin² θ

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

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