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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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If $ \cos^{-1}\large\frac{x}{a}$$+\cos^{-1}\large\frac{y}{b}= \alpha $, prove that $\large \frac{x^2}{a^2}-\frac{2xy}{ab}$$\cos \alpha+\large\frac{y^2}{b^2}$$=\sin^2\alpha. $

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  • \( cos^{-1}x+cos^{-1}y=cos^{-1} \bigg[ xy-\sqrt{1-x^2}\sqrt{1-y^2} \bigg] \)
\( cos^{-1}\frac{x}{a}+cos^{-1}\frac{y}{b}=\alpha\)
\( \Rightarrow cos^{-1} \bigg[ \frac{xy}{ab}-\sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}} \bigg] = \alpha\)
squaring both sides and take \( cos^2\alpha=1-sin^2\alpha\)
we get the result
\( \frac{x^2y^2}{a^2b^2}+cos^2\alpha-\frac{2xy}{ab}cos\alpha=1+\frac{x^2y^2}{a^2b^2}= \frac{x^2}{a^2}-\frac{y^2}{b^2}\)
\( \Rightarrow \frac{x^2}{a^2}-\frac{2xy}{ab}cos \alpha-\frac{y^2}{b^2}=sin^2\alpha\)
answered Mar 1, 2013 by thanvigandhi_1
 

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