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

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  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\:\:xy<1 \)
  • \(tan^{-1}1=\large\frac{\pi}{4}\)
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.

 

answered Feb 20, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 
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