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Prove that $ \tan^{-1} \large\frac{1}{7}$$ + 2\tan^{-1} \large\frac{1}{3} = \frac{\pi}{4} $

1 Answer

Toolbox:
  • \( 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}\)
  • \(tan^{-1}x+tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg) xy< 1\)
L.H.S
\( = tan^{-1}\large\frac{1}{7}+tan^{-1}\large\frac{3}{4} \)
\( \Rightarrow tan^{-1}\bigg( \large\frac{\large\frac{1}{7}+\large\frac{3}{4}}{1-\large\frac{3}{28}}\bigg)\)
\(=tan^{-1}=\large\frac{\pi}{4}=R.H.S\)

 

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