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Prove that \[cot(\frac{\pi}{4}-2cot^{-1}3)=7\]

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Toolbox:
  • \( 2cot^{-1}x= cot^{-1} \bigg( \large\frac{x^2-1}{2x} \bigg)\)
  • \(cot\large\frac{\pi}{4}=1\)
  • \( cot(A-B)=\large\frac{cotAcotB+1}{cotB-CotA}\)
L.H.S
By taking x=3, \(\large\frac{x^2-1}{2x}=\large\frac{9-1}{6}=\large\frac{8}{6}=\large\frac{4}{3}\)
\( 2cot^{-1}3= cot^{-1} \bigg( \large\frac{3^2-1}{2.3} \bigg)=cot^{-1}\large\frac{4}{3}\)
 
By substituting \(2cot^{-1}3\) in L.H.S. we get
\( cot \bigg[ \large\frac{\pi}{4}-cot^{-1}\bigg( \large\frac{4}{3} \bigg) \bigg] \)
By taking A=\(\large\frac{\pi}{4}\:and\:B=cot^{-1}\large\frac{4}{3}\),we get
\(\large\frac{cotAcotB+1}{cotB-CotA}=\large\frac{cot\large\frac{\pi}{4}.cot(cot^{-1}\large\frac{4}{3})+1}{cot(cot^{-1}\large\frac{4}{3})-cot\large\frac{\pi}{4}}\)
 
From the above formula of Cot(A-B) we get
\(\Rightarrow\:\large \frac{\large\frac{4}{3}+1}{\large\frac{4}{3}-1}\)
\(=\large\frac{\large\frac{7}{3}}{\large\frac{1}{3}}=7\)
 
=R.H.S.

 

answered Mar 2, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 
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