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Show that $\cos\bigg(2\tan^{-1}{\frac{1}{7}}\bigg)=\sin\bigg(4\tan^{-1}{\frac{1}{3}}\bigg)$

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Toolbox:
  • \( 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2} \)
  • \( tan^{-1}x=cos^{-1}\large\frac{1}{\sqrt{1+x^2}}\)
  • \( tan^{-1}x=sin^{-1}\large\frac{x}{\sqrt{1+x^2}}\)
L.H.S.
By taking \(x=\large\frac{1}{7},\:v\frac{2x}{1-x^2}=\large\frac{2.\large\frac{1}{7}}{1-\large\frac{1}{49}}=\large\frac{2}{7}.\large\frac{49}{48}=\large\frac{7}{24}\)
 
By substituting in the formula of \(2tan^{-1}x\) we get
\(2tan^{-1}\large\frac{1}{7}=tan^{-1}\large\frac{7}{24}\)
By taking \(x=\large\frac{7}{24},\large\frac{1}{\sqrt{1+x^2}}=\large\frac{1}{\sqrt{1+\large\frac{49}{576}}}=\large\frac{24}{25}\)
 
By substituting in the formula of \(tan^{-1}x\), we get
\(tan^{-1}\large\frac{7}{24}=cos^{-1}\large\frac{24}{25}\)
\( cos \bigg[ 2tan^{-1}\large\frac{1}{7} \bigg] \)
\( =cos \bigg[ tan^{-1}\large\frac{7}{24} \bigg] = cos\big(cos^{-1}\large\frac{24}{25}\big)=\large\frac{24}{25} \)
 
R.H.S.=
By taking \(x=\large\frac{1}{3},\:\large\frac{2x}{1-x^2}=\large\frac{2.\large\frac{1}{3}}{1-\large\frac{1}{9}}=\large\frac{2}{3}.\large\frac{9}{8}=\large\frac{3}{4}\)
 
By substituting in the formula of \(2tan^{-1}x\) we get
\(2tan^{-1}\large\frac{1}{3}=tan^{-1}\large\frac{3}{4}\)
\(\Rightarrow\:2.2tan^{-1}\large\frac{1}{3}=2tan^{-1}\large\frac{3}{4}\)
 
Again by taking x=\(\large\frac{3}{4}\) we get
\(4tan^{-1}\large\frac{1}{3}=2\: tan^{-1}\large\frac{3}{4} =tan^{-1}\large\frac{24}{7}\)
 
By taking \(x=\large\frac{24}{7},\:\large\frac{x}{\sqrt{1+x^2}}=\large\frac{\large\frac{24}{7}}{\sqrt{1+\large\frac{576}{49}}}=\large\frac{24}{7}.\large\frac{7}{25}=\large\frac{24}{25}\)
\(tan^{-1}\large\frac{24}{7}=sin^{-1}\large\frac{24}{25}\)
\( sin \bigg( 4tan^{-1}\large\frac{1}{3} \bigg) =sin \bigg( 2.2tan^{-1}\large\frac{1}{3} \bigg)\)
\( = sin \bigg( 2\: tan^{-1}\large\frac{3}{4} \bigg) \)
\( = sin \bigg( tan^{-1}\large\frac{24}{7} \bigg) \)
\( = sin \: sin^{-1}\large\frac{24}{25}=\large\frac{24}{25} \)
 
\(\Rightarrow\:R.H.S.=L.H.S.\)

 

answered Feb 19, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 

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