Question

If \[\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = \theta,\]  then 9x² − 12xy cos θ + 4y² is equal to

Options

• 36

•  −36 sin² θ

• 36 sin² θ

• 36 cos² θ

Answer 1

(c) 36 sin² θ

We know
\[\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} \left[ xy - \sqrt{1 - x^2}\sqrt{1 - y^2} \right]\]
Now,
\[\cos^{- 1} \frac{x}{2} + \cos^{- 1} \frac{y}{3} = \theta\]
\[ \Rightarrow \cos^{- 1} \left[ \frac{x}{2}\frac{y}{3} - \sqrt{1 - \frac{x^2}{4}}\sqrt{1 - \frac{y^2}{3}} \right] = \theta\]
\[ \Rightarrow \frac{x}{2}\frac{y}{3} - \sqrt{1 - \frac{x^2}{4}}\sqrt{1 - \frac{y^2}{3}} = \cos\theta\]
\[ \Rightarrow xy - \sqrt{4 - x^2}\sqrt{9 - y^2} = 6\cos\theta\]
\[ \Rightarrow \sqrt{4 - x^2}\sqrt{9 - y^2} = xy - 6\cos\theta\]
\[ \Rightarrow \left( 4 - x^2 \right)\left( 9 - y^2 \right) = x^2 y^2 + 36 \cos^2 \theta - 12xy\cos\theta (\text{ Squaring both the sides })\]
\[ \Rightarrow 36 - 4 y^2 - 9 x^2 + x^2 y^2 = x^2 y^2 + 36 \cos^2 \theta - 12xy\cos\theta\]
\[ \Rightarrow 36 - 4 y^2 - 9 x^2 = 36 \cos^2 \theta - 12xy\cos\theta\]
\[ \Rightarrow 9 x^2 - 12xy\cos\theta + 4 y^2 = 36 - 36 \cos^2 \theta\]
\[ \Rightarrow 9 x^2 - 12xy\cos\theta + 4 y^2 = 36 \sin^2 \theta\]