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# Considering the principal solutions, find the number of solutions of $\tan^{-1}2x+\tan^{-1}3x=\frac{\pi}{4}$

Toolbox:
• $tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\:\:xy<1$
• $tan^{-1}1=\large\frac{\pi}{4}$
By taking 2x in the place of x and 3x in the place of y in the above formula we get
$tan^{-1}2x+tan^{-1}{3x}= tan^{-1}\large\frac{2x+3x}{1-6x^2}=tan^{-1}\large\frac{5x}{1-6x^2}$

Substituting the value in the given eqn. we get
$tan^{-1}\large\frac{5x}{1-6x^2}=\large\frac{\pi}{4}=tan^{-1}1$
$\Rightarrow\:\large\frac{5x}{1-6x^2}=1$
$\Rightarrow 5x=1-6x^2$
$\Rightarrow \:6x^2+5x-1=0$

solving which $x=-1 \: or \: \large\frac{1}{6}$
2 solutions are possible.

edited Mar 19, 2013