Question

$\mathrm{If}{\cos }^{-1}\frac{x}{3}+{\cos }^{-1}\frac{y}{2}=\frac{\mathrm{\theta }}{2},\mathrm{then}4x^2-12xy\cos \frac{\mathrm{\theta }}{2}+9y^2=$
(a) 36
(b) 36 − 36 cos θ
(c) 18 − 18 cos θ
(d) 18 + 18 cos θ

Answer 1

(c) 18 − 18 cosθ

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

$$\begin{gathered} \therefore {\cos }^{-1}\frac{x}{3}+{\cos }^{-1}\frac{y}{2}=\frac{\theta }{2} \\ \Rightarrow {\cos }^{-1}\left(\frac{x}{3}\frac{y}{2}-\sqrt{1-\frac{x^2}{9}}\sqrt{1-\frac{y^2}{4}}\right)=\frac{\theta }{2} \\ \Rightarrow \frac{xy}{6}-\sqrt{\frac{9-x^2}{9}}\sqrt{\frac{4-y^2}{4}}=\cos \frac{\theta }{2} \\ \Rightarrow xy-6\cos \frac{\theta }{2}=\sqrt{9-x^2}\sqrt{4-y^2} \end{gathered}$$

Squaring both the sides, we get
$$\begin{gathered} x^2{y}^2-12xy\cos \frac{\theta }{2}+36{\cos }^2\frac{\theta }{2}=\left(9-x^2\right)\left(4-y^2\right) \\ \Rightarrow x^2{y}^2-12xy\cos \frac{\theta }{2}+36{\cos }^2\frac{\theta }{2}=36-9y^2-4x^2+x^2{y}^2 \\ \Rightarrow 4x^2+9y^2-12xy{\cos }^2\frac{\theta }{2}=36-36{\cos }^2\frac{\theta }{2} \\ \Rightarrow 4x^2+9y^2-12xy{\cos }^2\frac{\theta }{2}=36\left\{1-\left(\frac{\cos \theta +1}{2}\right)\right\}\left[\because \cos 2x=2{\cos }^{2}x-1\right] \\ \Rightarrow 4x^2+9y^2-12xy{\cos }^2\frac{\theta }{2}=18-18\cos \theta \end{gathered}$$