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If $ \cos^{-1}\frac{x}{a}+\cos^{-1}\frac{y}{b}=\alpha,$prove that $\frac{x^2}{a^2}-\frac{2xy}{ab}\cos \alpha+\frac{y^2}{b^2}=\sin^2\alpha.$

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Toolbox:
  • \( cos^{-1}x+cos^{-1}y=cos^{-1} \bigg[ xy+\sqrt{1-x^2}. \sqrt{1-y^2} \bigg] \)
  • \(cos^2\alpha=1-sin^\alpha\)
Given \( cos^{-1}\large\frac{x}{a}+cos^{-1}\large\frac{y}{b}=\alpha\)
 
\(Taking\:\large\frac{x}{a}\:in\:the\:place\:of\:x\:and\:\)
\(\large\frac{y}{a}\:in\:the\:place\:of\:y\:in\:the\:above\:formula\)
 
\( \Rightarrow cos^{-1} \bigg[ \large\frac{xy}{ab}-\sqrt{1-\large\frac{x^2}{a^2}}\sqrt{1-\large\frac{y^2}{b^2}} \bigg] = \alpha \)
\( \Rightarrow cos\alpha = \large\frac{xy}{ab}-\sqrt{1-\large\frac{x^2}{a^2}}\sqrt{1-\large\frac{y^2}{b^2}} \)
\(\Rightarrow cos\alpha - \large\frac{xy}{ab}=-\sqrt{1-\large\frac{x^2}{a^2}}\sqrt{1-\large\frac{y^2}{b^2}} \)
 
squaring both sides and write \(cos^2\alpha = 1-sin^2\alpha \)
\(\Rightarrow\:cos^2\alpha+(\frac{xy}{ab})^2-2\large\frac{xy}{ab}cos\alpha=\)
\(\bigg(1-\large\frac{x^2}{a^2}\bigg)\bigg(1-\large\frac{y^2}{b^2}\bigg)=1+(\large\frac{xy}{ab})^2-\large\frac{x^2}{a^2}-\large\frac{y^2}{b^2}\)
 
\(\Rightarrow\:\large\frac{x^2}{a^2}+\large\frac{y^2}{b^2}-2\large\frac{xy}{ab}=sin^2\alpha\)

 

answered Feb 20, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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