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Prove that $ 2\tan^{-1} \large\frac{1}{5}$$ + \tan^{-1} \large\frac{1}{8}$$ = \tan^{-1}\large \frac{4}{7} $

1 Answer

Toolbox:
  • \( 2tan^{-1}x=tan^{-1}\bigg[ \large\frac{2x}{1-x^2}\bigg]\)
  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy} xy < 1\)
L.H.S
\( tan^{-1} \bigg[ \large\frac{\large\frac{2}{5}}{1-\large\frac{1}{25}} \bigg] + tan^{-1}\large\frac{1}{8}\)
\( = tan^{-1} \bigg( \large\frac{5}{12} \bigg) + tan^{-1}\large\frac{1}{8}\)
\( = tan^{-1} \bigg(\large \frac{\large\frac{5}{12}+\large\frac{1}{8}}{1-\large\frac{5}{96}} \bigg)\)
\( = tan^{-1} \large\frac{4}{7}= R.H.S \)

 

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

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