Question

lf ${\cos }^{-1}\frac{x}{2}+{\cos }^{-1}\frac{y}{3}=\frac{\pi }{6}$, then value of $\frac{x^2}{4}-\frac{xy}{2\sqrt{3}}+\frac{y^2}{9}$ is :

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{3}{2}$

Answer 1

The correct option is C $\frac{1}{4}$

$co{s}^{-1}\left(\frac{x}{2}\right)+co{s}^{-1}\left(\frac{y}{3}\right)=\frac{\pi }{6}$
$co{s}^{-1}\frac{y}{3}=\frac{\pi }{6}-co{s}^{-1}\frac{x}{2}$
take cosine on both the sides, we have

$\frac{y}{3}=\frac{\sqrt{3}}{2}\times \frac{x}{2}+\frac{1}{2}\sqrt{1-\frac{x^2}{4}}$

$(cosco{s}^{-1}x=x,cos(A-B)=cosAcosB-sinAsinB)$
$\frac{y}{3}=\frac{\sqrt{3}x}{4}+\frac{\sqrt{4-x^2}}{4}$
$(\frac{4y}{3}-\sqrt{3}x)=\sqrt{4-x^2}$
$\frac{16y^2}{9}+3x^2-\frac{8}{\sqrt{3}}xy=4-x^2$

$4x^2+\frac{16y^2}{9}-\frac{8xy}{\sqrt{3}}=4$
$\Rightarrow \frac{x^2}{4}-\frac{xy}{\sqrt{3}}+\frac{y^2}{9}=\frac{4}{16}=\frac{1}{4}$