# If $\cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2}$, then $4 x^{2}-12 x y \cos \frac{\theta}{2}+9 y^{2}=$

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

Solution:
(c) 18 − 18 cosθ
We know
$\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)$
$\therefore \cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2}$
$\Rightarrow \cos ^{-1}\left(\frac{x}{3} \frac{y}{2}-\sqrt{1-\frac{x^{2}}{9}} \sqrt{1-\frac{y^{2}}{4}}\right)=\frac{\theta}{2}$
$\Rightarrow \frac{x y}{6}-\sqrt{\frac{9-x^{2}}{9}} \sqrt{\frac{4-y^{2}}{4}}=\cos \frac{\theta}{2}$
$\Rightarrow x y-6 \cos \frac{\theta}{2}=\sqrt{9-x^{2}} \sqrt{4-y^{2}}$
Squaring both the sides, we get
$x^{2} y^{2}-12 x y \cos \frac{\theta}{2}+36 \cos ^{2} \frac{\theta}{2}=\left(9-x^{2}\right)\left(4-y^{2}\right)$
$\Rightarrow x^{2} y^{2}-12 x y \cos \frac{\theta}{2}+36 \cos ^{2} \frac{\theta}{2}=36-9 y^{2}-4 x^{2}+x^{2} y^{2}$
$\Rightarrow 4 x^{2}+9 y^{2}-12 x y \cos ^{2} \frac{\theta}{2}=36-36 \cos ^{2} \frac{\theta}{2}$
$\Rightarrow 4 x^{2}+9 y^{2}-12 x y \cos ^{2} \frac{\theta}{2}=36\left\{1-\left(\frac{\cos \theta+1}{2}\right)\right\} \quad\left[\because \cos 2 x=2 \cos ^{2} x-1\right]$
$\Rightarrow 4 x^{2}+9 y^{2}-12 x y \cos ^{2} \frac{\theta}{2}=18-18 \cos \theta$