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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve for x : $ \tan^{-1} \bigg(\large \frac{x-1}{x+1} \bigg) + \tan^{-1} \bigg( \frac{2x-1}{2x+1} \bigg) = \tan^{-1} \bigg( \frac{23}{36} \bigg). $

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  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy} , xy < 1\)
\( tan^{-1}\large\frac{x-1}{x+1}+tan^{-1}\large\frac{2x-1}{2x+1}=tan^{-1}\large\frac{23}{36}\)
\( \Rightarrow tan^{-1} \bigg[ \large\frac{\large\frac{x-1}{x+1}+\large\frac{2x-1}{2x+1}}{1-\large\frac{x-1}{x+1}.\large\frac{2x-1}{2x+1}} \bigg] = tan^{-1} \large\frac{23}{36}\)
 
Simplifying which
\( tan^{-1} \large\frac{2x^2-1}{3x}=\large\frac{23}{36}\)
\( \Rightarrow \large\frac{2x^2-1}{3x}=\large\frac{23}{36}\)
\( \Rightarrow 24x^2-23x-12=0\)
\( x = \large\frac{4}{3} \: or \: \large\frac{-3}{8}\)
 
But \( x=\large\frac{-3}{8}\) is not satisfied so,
\( x=\large\frac{4}{3}\)

 

answered Mar 1, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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