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प्रश्न
If \[\cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha, then\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = \]
पर्याय
sin2 α
cos2 α
tan2 α
cot2 α
उत्तर
(a) sin2 α
We know that
\[\cos^{- 1} x + \cos^{- 1} y = \cos^{- 1} \left( xy - \sqrt{1 - x^2}\sqrt{1 - y^2} \right)\]
\[\therefore \cos^{- 1} \frac{x}{a} + \cos^{- 1} \frac{y}{b} = \alpha\]
\[ \Rightarrow \cos^{- 1} \left( \frac{x}{a}\frac{y}{b} - \sqrt{1 - \frac{x^2}{a^2}}\sqrt{1 - \frac{y^2}{b^2}} \right) = \alpha\]
\[ \Rightarrow \frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}}\sqrt{1 - \frac{y^2}{b^2}} = \cos\alpha\]
\[ \Rightarrow \sqrt{1 - \frac{x^2}{a^2}}\sqrt{1 - \frac{y^2}{b^2}} = \frac{xy}{ab} - \cos\alpha\]
\[ \Rightarrow \left( 1 - \frac{x^2}{a^2} \right)\left( 1 - \frac{y^2}{b^2} \right) = \frac{x^2}{a^2}\frac{y^2}{b^2} + \cos^2 \alpha - \frac{2xy}{ab}\cos\alpha \left[\text{ Squaring both the sides }\right]\]
\[ \Rightarrow 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2}{a^2}\frac{y^2}{b^2} = \frac{x^2}{a^2}\frac{y^2}{b^2} + \cos^2 \alpha - \frac{2xy}{ab}\cos\alpha\]
\[ \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab}\cos\alpha = 1 - \cos^2 \alpha = \sin^2 \alpha\]
\[\]
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