Question

If  \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]

Options

• sin² α

• cos² α

• tan² α

• cot² α

Answer 1

(a) sin² α
We know that
\[\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} \left( xy - \sqrt{1 - x^2}\sqrt{1 - y^2} \right)\]
\[\therefore \cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha\]
\[ \Rightarrow \cos^{- 1} \left( \frac{x}{a}\frac{y}{b} - \sqrt{1 - \frac{x^2}{a^2}}\sqrt{1 - \frac{y^2}{b^2}} \right) = \alpha\]
\[ \Rightarrow \frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}}\sqrt{1 - \frac{y^2}{b^2}} = \cos\alpha\]
\[ \Rightarrow \sqrt{1 - \frac{x^2}{a^2}}\sqrt{1 - \frac{y^2}{b^2}} = \frac{xy}{ab} - \cos\alpha\]
\[ \Rightarrow \left( 1 - \frac{x^2}{a^2} \right)\left( 1 - \frac{y^2}{b^2} \right) = \frac{x^2}{a^2}\frac{y^2}{b^2} + \cos^2 \alpha - \frac{2xy}{ab}\cos\alpha \left[\text{ Squaring both the sides }\right]\]
\[ \Rightarrow 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2}{a^2}\frac{y^2}{b^2} = \frac{x^2}{a^2}\frac{y^2}{b^2} + \cos^2 \alpha - \frac{2xy}{ab}\cos\alpha\]
\[ \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab}\cos\alpha = 1 - \cos^2 \alpha = \sin^2 \alpha\]
\[\]