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Show that $ \sin^{-1}\large\frac{12}{13}$$+\cos^{-1}\large\frac{4}{5}$$+\tan^{-1}\large\frac{63}{16} $$= \pi $

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Toolbox:
  • \( sin^{-1}\large\frac{12}{13}=tan^{-1}\large\frac{12}{5}\)
  • \( cos^{-1}\large\frac{4}{5}=tan^{-1}\large\frac{3}{4}\)
  • \( tan^[-1}x+tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y]{1-xy} \bigg)
  • \( xy > 1\)
\( sin^{-1}\large\frac{12}{13}+cos^{-1}\large\frac{4}{5}+tan^{-1}\large\frac{63}{16}\)
\( = tan^{-1}\large\frac{12}{5}+tan^{-1}\large\frac{3}{4}+tan6{-1}\large\frac{63}{13}\)
\( = \pi+tan^{-1} \bigg[ \large\frac{\large\frac{12}{5}+\large\frac{3}{4}}{1-\large\frac{36}{20}} \bigg] +tan^{-1}\large\frac{63}{13}\)
\( = \pi+tan^{-1}\large\frac{63}{-13}+tan^{-1}\large\frac{63}{13}\)
\( = \pi = R.H.S\)

 

answered Feb 28, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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