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Prove that \[ 2sin^{-1} \frac{3}{5} = tan^{-1} \frac{24}{7} \]

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Toolbox:
  • \( sin^{-1}x=tan^{-1}\large\frac{x}{\sqrt{1-x^2}}\)
  • \( 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2} |x| < 1\)
Given $2sin^{-1}\large \frac{3}{5} = tan^{-1} \large\frac{24}{7}$
 
L.H.S.: Let $x =$ \(\large\frac{3}{5}\), \( \rightarrow \large \frac{x}{\sqrt{1-x^2}}=\large\frac{\large\frac{3}{5}}{\sqrt{1-\large\frac{9}{25}}}=\large\frac{3}{4}\)
 
\(\Rightarrow\:sin^{-1}\large\frac{3}{5}=tan^{-1}\large\frac{3}{4}\)
\(\Rightarrow 2sin^{-1}\large\frac{3}{5}=2tan^{-1}\large\frac{3}{4}\)
Given \( 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}\), we need to evaluate \(2tan^{-1}\large\frac{3}{4}\)
 
Let $x =\large\frac{3}{4} \rightarrow \large \large\frac{2x}{1-x^2}=\large\frac{2.\frac{3}{4}}{1-\large\frac{9}{16}}=\large\frac{6}{4}.\large\frac{16}{7}=\large\frac{24}{7}$
 
\( \Rightarrow\:2tan^{-1}\large\frac{3}{4} = tan^{-1}\large\frac{24}{7} =\) R.H.S.
answered Jul 9, 2014 by balaji.thirumalai
 
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