Question

lf ${\sin }^{-1}\frac{x}{3}+{\sin }^{-1}\frac{y}{4}=\frac{\pi }{6}$ , then value of $\frac{x^2}{9}+\frac{xy}{4\sqrt{3}}+\frac{y^2}{16}$ is

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

Answer 1

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

$si{n}^{-1}\frac{x}{3}=\frac{\pi }{6}-si{n}^{-1}\left(\frac{y}{4}\right)$
take sin on both sides

$\frac{x}{3}=sin\frac{\pi }{6}\sqrt{1-\frac{y^2}{16}}-cos\frac{\pi }{6}\frac{y}{4}[sinsi{n}^{-1}x=x,sin(A-B)=sinAcosB-sinBcosA]$

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

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

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