Question

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

• A. $36$
• B. $-36{\sin }^2\theta$
• C. $36{\sin }^2\theta$
• D. $36{\cos }^2\theta$

Answer 1

The correct option is C $36{\sin }^2\theta$

Given, ${\cos }^{-1}\frac{x}{2}+{\cos }^{-1}\frac{y}{3}=\theta$

$\Rightarrow {\cos }^{-1}\{\frac{xy}{6}-\sqrt{1-\frac{x^2}{4}}.\sqrt{1-\frac{y^2}{9}}\}=\theta$

$\Rightarrow xy-\sqrt{4-x^2}.\sqrt{9-y^2}=6\cos \theta$

$\Rightarrow (xy-6\cos \theta {)}^2=(4-x^2\left)\right(9-y^2)$

$\Rightarrow -12xy\cos \theta +36{\cos }^2\theta =36-4y^2-9x^2$

$\Rightarrow 9x^2+4y^2-12xy\cos \theta =36{\sin }^2\theta$