If ^ -1 x 2 + ^ -1 y 2 = , then 9x^2-12xy +4y^2 is equal to:

MCQ

If $\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{2}=\theta,$ then $\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2$ is equal to:

A
$36$
B
$-36\sin^2\theta$
✓
$36\sin^2\theta$
D
$-36\cos^2\theta$

✓

Answer

Correct option: C.

$36\sin^2\theta$

We know
$\cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big[\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big]$
Now,
$\cos^{-1}\frac{\text{x}}{2}+\cos^{-1}\frac{\text{y}}{2}=\theta$
$\Rightarrow\cos^{-1}\Big[\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}\Big]=\theta$
$\Rightarrow\frac{\text{x}}{2}\frac{\text{y}}{3}-\sqrt{1-\frac{\text{x}^2}{4}}\sqrt{1-\frac{\text{y}^2}{3}}=\cos\theta$
$\Rightarrow\text{xy}-\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=6\cos\theta$
$\Rightarrow\sqrt{4-\text{x}^2}\sqrt{9-\text{y}^2}=\text{xy}-6\cos\theta$
$\Rightarrow(4-\text{x}^2)(9-\text{y}^2)=\text{x}^2\text{y}^2+36\cos^2\theta-12\text{xy}\cos\theta ($Squaring both the sides$)$
$\Rightarrow36-4\text{y}^2-9\text{x}^2+\text{x}^2\text{y}^2=\text{x}^2\text{y}^2+36\cos^{2}\theta-12\text{xy}\cos\theta$
$\Rightarrow36-4\text{y}^2-9\text{x}^2=36\cos^2\theta-12\text{xy}\cos\theta$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36-36\cos^2\theta$
$\Rightarrow9\text{x}^2-12\text{xy}\cos\theta+4\text{y}^2=36\sin^2\theta$

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