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2014-12-18T22:25:28+05:30

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Given problem is incorrect.  The correct one is to solve for x given that : sum (not difference) is pi/4.

tan^{-1}2x+tan^{-1}3x=\pi/4\\\\tan(tan^{-1}2x+tan^{-1}3x)=tan\ \pi/4\\\\\frac{2x+3x}{1-2x*3x}=1\\\\6x^2+5x-1=0\\\\x=\frac{-5+-\sqrt{25+24}}{12}=-1\ or\ 1/6\\\\

Formula :

   tan^{-1}a+tan^{-1}b\\.\ \ \ \ =\ tan^{-1}(\frac{x+y}{1-xy}),\ \ \ xy<1\\.\ \ \ \ =\pi-tan^{-1}(\frac{x+y}{1-xy}),\ \ xy>1\\.\ \ \ \ =\pi/2,\ \ \ xy=1\\

In the above example, in both cases,  6x^2 < 1 -  solution is  x = 1/6 and  for 6x^2 > 1,  we have the solution as x = -1.

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