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What does $sin^{-1} \frac{5}{13}+cos^{-1} \frac {3}{5}$ reduce to?

$\begin{array}{1 1} tan^{-1}\frac{63}{16} \\ sin^{-1}\frac{63}{16} \\cos^{-1}\frac{63}{16} \\ cot^{-1}\frac{63}{16} \end{array} $

Can you answer this question?
 
 

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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 $sin^{-1} \large\frac{5}{13}+cos^{-1} \large\frac {3}{5}$
We know that \( sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}\)
 
By taking $x=\large\frac{5}{13} \rightarrow$ \(\large \frac{x}{\sqrt{1-x^2}}\)=\(\Large \frac{\frac{5}{13}}{\sqrt{1-\large\frac{25}{169}}}\)\(=\large\frac{5}{13}.\large\frac{13}{12}=\large\frac{5}{12}\)
\(\Rightarrow\:sin^{-1}\large\frac{5}{13}=tan^{-1}\large\frac{5}{12}\)
We know that \( cos^{-1}x=tan^{-1}\large \frac{\sqrt{1-x^2}}{x}\)
 
By taking $x=$\(\frac{3}{5} \rightarrow \large \frac{\sqrt{1-x^2}}{x}=\Large \frac{\sqrt{1-\large\frac{9}{25}}}{\large\frac{3}{5}}\)\(=\large\frac{4}{5}.\large\frac{5}{3}=\large\frac{3}{4}\)
 
\(\Rightarrow\:cos^{-1}\large\frac{3}{5}=tan^{-1}\large\frac{4}{3}\)
\( \Rightarrow sin^{-1} \large\frac{5}{13}+cos^{-1}\large\frac{3}{5}\) \( = tan^{-1}\large\frac{5}{12}+tan^{-1}\large\frac{4}{3}\)
 
 
\(\large\frac{x+y}{1-xy}=\large\frac{\large\frac{5}{12}+\large\frac{4}{3}}{1-\large\frac{5}{12}.\large\frac{4}{3}}=\large\frac{63}{36}.\large\frac{36}{16}=\large\frac{63}{16}\)
\( = tan^{-1} \bigg[ \large\frac{\large\frac{5}{12}+\large\frac{4}{3}}{1-\large\frac{20}{36}} \bigg] \)
\( = tan^{-1}\large\frac{63}{16} \)
=L.H.S
 
We know that \(tan^{-1}x+tan{-1}y=tan^{-1}\large \frac{x+y}{1-xy}\)
 
By taking \(x=\large\frac{5}{12}\:and\:y=\large\frac{4}{3}\:in\:tan^{-1}x+tan^{-1}y\)
$x+y = \large\frac{5}{12}+\large\frac{4}{3} = \large\frac{63}{36}$
$1 - xy = 1 - \large\frac{5}{12} \times \frac{4}{3} = \large\frac{16}{36}$
$\large \frac{x+y}{1-xy} =\large \frac{63}{36}.\large\frac{36}{16}=\large\frac{63}{16}$
 
$\Rightarrow tan^{-1}\large\frac{5}{12}+tan^{-1}\large\frac{4}{3} =$ \( tan^{-1}\large\frac{63}{16} \)

 

answered Feb 23, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 
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