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Solve : $ \tan^{-1}2x+tan^{-1}3x = \large\frac{\pi}{4}. $

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  • \( tan^{-1}x+tan^{-1}y=tan^{-1} \bigg(\large \frac{x+y}{1-xy} \bigg)\)
\( tan^{-1}\large\frac{5x}{1-6x^2}=\frac{\pi}{4}\)
\( \Rightarrow 5x=1-6x^2\)
\( \Rightarrow x=\large\frac{1}{6}, -1\)

 

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

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