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Prove the following: \[ 2 \: tan^{-1} \frac{1}{2} + tan ^{-1} \frac{1}{7} = tan^{-1} \frac{31}{17} \]

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Toolbox:
  • \( 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2} for\:|x|<1\)
  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\: \: \: xy<1\)
L.H.S:
\(\large\frac{1}{2}\:is\:<1\:so\:2tan^{-1}\large\frac{1}{2}=tan^{-1} \bigg( \large\frac{1}{1-\frac{1}{4}} \bigg) \)
=\(tan^{-1}\large\frac{4}{3}\)
\(2tan^{-1}\large\frac{1}{2}+tan^{-1}\large\frac{1}{7}=tan^{-1}\large\frac{4}{3}+tan^{-1}\large\frac{1}{7}\)
 
By taking \(x=\large\frac{4}{3}\:and\:y=\large\frac{1}{7} \)
\(\:and\:since \large\frac{4}{3}.\large\frac{1}{7}=\large\frac{4}{21}\:is\:<1\) by using the above formula of \(tan^{-1}x+tan^{-1}y\)
 
\(\Rightarrow\:L.H.S.= tan^{-1} \bigg[ \large\frac{\large\frac{4}{3}+\large\frac{1}{7}}{1-\large\frac{4}{21}} \bigg]\)
\( = tan^{-1}\frac{31}{17}=R.H.S\)

 

answered Mar 2, 2013 by rvidyagovindarajan_1
edited Mar 15, 2013 by thanvigandhi_1
 

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