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

Solution:
(c) $36 \sin ^{2} \theta$
We know
$\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left[x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right]$
$\Rightarrow \cos ^{-1}\left[\frac{x}{2} \frac{y}{3}-\sqrt{1-\frac{x^{2}}{4}} \sqrt{1-\frac{y^{2}} {3}}\right]=\theta$
$\Rightarrow \frac{x}{2} \frac{y}{3}-\sqrt{1-\frac{x^{2}}{4}} \sqrt{1-\frac{y^{2}}{3}}=\cos \theta$
$\Rightarrow x y-\sqrt{4-x^{2}} \sqrt{9-y^{2}}=6 \cos \theta$
$\Rightarrow \sqrt{4-x^{2}} \sqrt{9-y^{2}}=x y-6 \cos \theta$
$\Rightarrow\left(4-x^{2}\right)\left(9-y^{2}\right)=x^{2} y^{2}+36 \cos ^{2} \theta-12 x y \cos \theta$ (Squaring both the sides)
$\Rightarrow 36-4 y^{2}-9 x^{2}+x^{2} y^{2}=x^{2} y^{2}+36 \cos ^{2} \theta-12 x y \cos \theta$
$\Rightarrow 36-4 y^{2}-9 x^{2}=36 \cos ^{2} \theta-12 x y \cos \theta$
$\Rightarrow 9 x^{2}-12 x y \cos \theta+4 y^{2}=36-36 \cos ^{2} \theta$
$\Rightarrow 9 x^{2}-12 x y \cos \theta+4 y^{2}=36 \sin ^{2} \theta$
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