Question

\[\text{ If }\cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}, \text{ then }4 x^2 - 12xy \cos\frac{\theta}{2} + 9 y^2 =\]

विकल्प

• 36

• 36 − 36 cos θ

• 18 − 18 cos θ

• 18 + 18 cos θ

Answer 1

(c) 18 − 18 cosθ

We know
\[\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} \left( xy - \sqrt{1 - x^2}\sqrt{1 - y^2} \right)\]
\[\therefore \cos^{- 1} \frac{x}{3} + \cos^{- 1} \frac{y}{2} = \frac{\theta}{2}\]
\[ \Rightarrow \cos^{- 1} \left( \frac{x}{3}\frac{y}{2} - \sqrt{1 - \frac{x^2}{9}}\sqrt{1 - \frac{y^2}{4}} \right) = \frac{\theta}{2}\]
\[ \Rightarrow \frac{xy}{6} - \sqrt{\frac{9 - x^2}{9}}\sqrt{\frac{4 - y^2}{4}} = \cos\frac{\theta}{2}\]
\[ \Rightarrow xy - 6\cos\frac{\theta}{2} = \sqrt{9 - x^2}\sqrt{4 - y^2}\]
Squaring both the sides, we get
\[x^2 y^2 - 12xy\cos\frac{\theta}{2} + 36 \cos^2 \frac{\theta}{2} = \left( 9 - x^2 \right)\left( 4 - y^2 \right)\]
\[ \Rightarrow x^2 y^2 - 12xy\cos\frac{\theta}{2} + 36 \cos^2 \frac{\theta}{2} = 36 - 9 y^2 - 4 x^2 + x^2 y^2 \]
\[ \Rightarrow 4 x^2 + 9 y^2 - 12xy \cos^2 \frac{\theta}{2} = 36 - 36 \cos^2 \frac{\theta}{2}\]
\[ \Rightarrow 4 x^2 + 9 y^2 - 12xy \cos^2 \frac{\theta}{2} = 36\left\{ 1 - \left( \frac{\cos\theta + 1}{2} \right) \right\} \left[ \because \cos2x = 2 \cos^2 x - 1 \right]\]
\[ \Rightarrow 4 x^2 + 9 y^2 - 12xy \cos^2 \frac{\theta}{2} = 18 - 18\cos\theta\]