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Show that $\sin^{-1}{\frac{5}{13}}+\cos^{-1}{\frac{3}{5}}=\tan^{-1}{\frac{63}{16}}$

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Toolbox:
  • \( sin^{-1}x=tan^{-1}\bigg( \frac{x}{\sqrt{1-x^2}} \bigg) \)
  • \( cos^{-1}x=tan^{-1} \bigg( \frac{\sqrt{1-x^2}}{x} \bigg) \)
  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy},\:\:\:xy<1 \)
L.H.S.=
By taking \(x=\frac{5}{13},\:\frac{x}{\sqrt{1-x^2}}=\frac{\frac{5}{13}}{\sqrt{1-\frac{25}{169}}}=\frac{5}{13}.\frac{13}{12}=\frac{5}{12}\)
By substituting in the above formula of \(sin^{-1}x\) we get
\( sin^{-1}\frac{5}{13}=tan^{-1}\frac{5}{12}\)
By taking \(x=\frac{3}{5},\:\frac{\sqrt{1-x^2}}{x}=\frac{\sqrt{1-\frac{9}{25}}}{\frac{3}{5}}=\frac{4}{5}.\frac{5}{3}=\frac{4}{3}\)
By substituting in the above formula of \(cos^{-1}x\) we get
\( cos^{-1}\frac{3}{5}=tan^{-1}\frac{4}{3} \)
By substituting the values, the given expression becomes
\(sin^{-1}\frac{5}{13}+cos^{-1}\frac{3}{5}=tan^{-1}\frac{5}{12}+tan^{-1}\frac{4}{3}\)
By taking \(x=\frac{5}{12},\:and\:y=\frac{4}{3}\)
\(\frac{x+y}{1-xy}=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}.\frac{4}{3}}=\frac{63}{36}.\frac{36}{16}=\frac{63}{16}\)
\(\Rightarrow\:tan^{-1}\bigg(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}.\frac{4}{3}}\bigg)=tan^{-1}\frac{63}{16}\)
=R.H.S.

 

answered Feb 19, 2013 by thanvigandhi_1
edited Mar 7, 2013 by rvidyagovindarajan_1
 

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