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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Prove that \( \tan^{-1} \large\frac{63}{16}=\sin^{-1}\bigg( \large\frac{5}{13} \bigg) + \cos^{-1} \bigg( \large\frac{3}{5} \bigg). \)

This is Q.No.7 of Misc. chapter 2

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Toolbox:
  • \( sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}\)
  • \( cos^{-1}x=tan^{-1}\large \frac{\sqrt{1-x^2}}{x}\)
  • \(tan^{-1}x+tan{-1}y=tan^{-1}\large \frac{x+y}{1-xy}\)
Given R.H.S. = $sin^{-1} \frac{5}{13}+cos^{-1} \frac {3}{5}$
We know that \( sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}\)
By taking $x=\frac{5}{13} \rightarrow$ \(\large \frac{x}{\sqrt{1-x^2}}\)=\(\Large \frac{\frac{5}{13}}{\sqrt{1-\frac{25}{169}}}\)\(=\frac{5}{13}.\frac{13}{12}=\frac{5}{12}\)
\(\Rightarrow\:sin^{-1}\frac{5}{13}=tan^{-1}\frac{5}{12}\)
We know that \( cos^{-1}x=tan^{-1}\large \frac{\sqrt{1-x^2}}{x}\)
By taking $x=$\(\frac{3}{5} \rightarrow \large \frac{\sqrt{1-x^2}}{x}=\Large \frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}\)\(=\frac{4}{5}.\frac{5}{3}=\frac{3}{4}\)
\(\Rightarrow\:cos^{-1}\frac{3}{5}=tan^{-1}\frac{4}{3}\)
\( \Rightarrow sin^{-1} \frac{5}{13}+cos^{-1}\frac{3}{5}\) \( = tan^{-1}\frac{5}{12}+tan^{-1}\frac{4}{3}\)
\(\frac{x+y}{1-xy}=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}.\frac{4}{3}}=\frac{63}{36}.\frac{36}{16}=\frac{63}{16}\)
\( = tan^{-1} \bigg[ \frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{20}{36}} \bigg] \)
\( = tan^{-1}\frac{63}{16} \)
=L.H.S
We know that \(tan^{-1}x+tan{-1}y=tan^{-1}\large \frac{x+y}{1-xy}\)
By taking \(x=\frac{5}{12}\:and\:y=\frac{4}{3}\:in\:tan^{-1}x+tan^{-1}y\)
$x+y = \frac{5}{12}+\frac{4}{3} = \frac{63}{36}$
$1 - xy = 1 - \frac{5}{12} \times \frac{4}{3} = \frac{16}{36}$
$\large \frac{x+y}{1-xy} = \frac{63}{36}.\frac{36}{16}=\frac{63}{16}$
$\Rightarrow tan^{-1}\frac{5}{12}+tan^{-1}\frac{4}{3} =$ \( tan^{-1}\frac{63}{16} \) = L.H.S.
answered Feb 28, 2013 by thanvigandhi_1
edited Mar 18, 2013 by rvidyagovindarajan_1
 

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