Question

If ${\cos }^{-1}\frac{x}{2}+{\cos }^{-1}\frac{y}{3}=\pi /6$ than the value of $\frac{x^2}{4}-\frac{xy}{2\sqrt{3}}+\frac{y^2}{9}$ is

• A. $3/4$
• B. $1/2$
• C. $1/4$
• D. $3/2$

Answer 1

The correct option is C $1/4$

$\to$ let ${\cos }^{-1}\frac{x}{2}=A,{\cos }^{-1}\frac{y}{3}=B$

As $cos(A+B)=\cos A,\cos B-\sin A,\sin B$

and $co{s}^{-1}x={\sin }^{-1}\sqrt{1-x^2}$

$\Rightarrow \cos (A+B)=\frac{x}{2}\times \frac{y}{3}-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{9}}$

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

$\Rightarrow (\frac{\sqrt{3}}{2}-\frac{xy}{6}{)}^2=(-\sqrt{1-\frac{x^2}{4}}\sqrt{1-\frac{y^2}{9}}{)}^2$

$\Rightarrow \frac{3}{4}+\frac{x^2{y}^2}{36}-\frac{\sqrt{3}xy}{6}=(1-\frac{x^2}{4})(1-\frac{y^2}{9})$

$\Rightarrow \frac{3}{4}+\frac{x^2{y}^2}{36}-\frac{xy}{2\sqrt{3}}=1+\frac{x^2}{4}-\frac{y^2}{9}+\frac{x^2{y}^2}{36}$

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

$\Rightarrow \left(C\right)$