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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Simplify\( 2\tan^{-1} \bigg(\large \frac{1}{5} \bigg) + \sec^{-1} \bigg( \large\frac{5\sqrt 2}{7} \bigg) + 2\: tan^{-1} \bigg( \frac{1}{8} \bigg) \)

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Toolbox:
  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\bigg(\large\frac{x+y}{1-xy} \bigg) xy<1\)
  • \(sec^{-1}x=tan^{-1}\sqrt{x^2-1}\)
  • \(2tan^{-1}x=tan^{-1} \bigg( \large\frac{2x}{1-x^2} \bigg)\)
\( 2 \bigg[ tan^{-1}\large\frac{1}{5}+tan^{-1}\large\frac{1}{8} \bigg]+tan^{-1} \sqrt{\large\frac{50}{49}-1}\)
\( 2\bigg[ tan^{-1} \bigg( \large\frac{\large\frac{1}{5}+\large\frac{1}{8}}{1-\large\frac{1}{40}} \bigg) \bigg]+tan^{-1}\large\frac{1}{7}\)
 
\( 2tan^{-1}\large\frac{1}{3}+tan^{-1}\large\frac{1}{7}\)
\( tan^{-1}\large\frac{3}{4}+tan^{-1}\large\frac{1}{7}\)
\( tan^{-1} \bigg( \large\frac{\large\frac{3}{4}=\large\frac{1}{7}}{1-\large\frac{3}{28}} \bigg) = tan^{-1}1\)
\( = \large\frac{\pi}{4}\)

 

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

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