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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Show that : $ \tan\large\frac{1}{2} \bigg[ \sin^{-1}\frac{2x}{1+x^2} + \cos^{-1}\large\frac{1-y^2}{1+y^2} \bigg] =\large \frac{x+y}{1-xy}$$, |x| < 1, y > 0 \: xy < 1 $

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Toolbox:
  • Put \( x= tan\alpha \: and \: y=tan \beta\)
  • \(\large \frac{2tan\alpha}{1+tan^2\alpha}=sin2\alpha\)
  • \( tan(\alpha+\beta)\)
  • \( \large\frac{1+tan^2\alpha}{1+tan^2\beta}=cos2\beta\)
  • \( = \large\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}\)
Put \( x=tan \alpha\: \: y=tan \beta\)
\( \alpha = tan^{-1}x\: \: \: \beta=tan^{-1}y\)
then L.H.S = \( tan\large\frac{1}{2} \bigg[ sin^{-1}sin2\alpha+cos^{-1}cos2\beta \bigg]\)
\( = tan (\alpha + \beta)\)
\( = \large\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}\)
\( = \large\frac{x+y}{1-xy}=R.H.S\)

 

answered Feb 28, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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