First slide

Properties of ITF

Question

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

Moderate

A

$\frac{3}{4}$
B

$\frac{1}{2}$
C

$\frac{1}{4}$
D

None of these

Solution

$W e h a v e \cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = \frac{π}{6}$
$\Rightarrow \cos^{- 1} [\frac{x y}{6} - \sqrt{1 - \frac{x^{2}}{4}} \sqrt{1 - \frac{y^{2}}{9}}] = \frac{π}{6} \Rightarrow \frac{x y}{6} - \sqrt{1 - \frac{x^{2}}{4}} \sqrt{1 - \frac{y^{2}}{9}} = \cos \frac{π}{6} = \frac{\sqrt{3}}{2} \Rightarrow \frac{x y}{6} - \frac{\sqrt{3}}{2} = \sqrt{1 - \frac{x^{2}}{4}} \sqrt{1 - \frac{y^{2}}{9}} \Rightarrow {(\frac{x y}{6} - \frac{\sqrt{3}}{2})}^{2} = (1 - \frac{x^{2}}{4}) (1 - \frac{y^{2}}{9}) \Rightarrow \frac{x^{2}}{4} - \frac{x y}{2 \sqrt{3}} + \frac{y^{2}}{9} = \frac{1}{4}$

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