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Show that $2tan^{-1}(-3)=\frac{-\pi}{2}+tan^{-1}\bigg(\frac{-4}{3}\bigg)$

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Toolbox:
  • \( 2tan^{-1}x=-\pi +tan^{-1}\large\frac{2x}{1-x^2} \: if \: x < -1\)
  • \(tan^{-1}(-x)=-tan^{-1}x\)
  • \( tan^{-1}x+tan^{-1}y=tan^{-1} \bigg(\large \frac{x+y}{1-xy} \bigg)\:\:if\:xy<1\)
  • \(tan^{-1}\infty=\large\frac{\pi}{2}\)
The question can be rewritten as prove \( 2tan^{-1}(-3)-tan^{-1} \bigg(\large\frac{-4}{3}\bigg)=-\large\frac{\pi}{2}\)
By taking x=-3 we get \(\large\frac{2x}{1-x^2}=\frac{2.-3}{1-9}=\large\frac{6}{8}=\large\frac{3}{4}\)
Substituting in the above formula \(2tan^{-1}(-3)=-\pi+tan^{-1}\large\frac{3}{4}\)
\(tan^{-1}(-\frac{4}{3})=-tan^{-1}\large\frac{4}{3}\)
 
L.H.S.=
\( 2tan^{-1}(-3)-tan^{-1} \bigg(\large\frac{-4}{3}\bigg)\)
\(= -\pi+tan^{-1}\bigg( \large\frac{3}{4} \bigg)-tan^{-1} \bigg( \large\frac{-4}{3}\bigg)\)
\( = -\pi+tan^{-1}\large\frac{3}{4}+tan^{-1}\large\frac{4}{3}\)
By taking \(x=\large\frac{3}{4}\:and\:y=\large\frac{4}{3}\),
\(\large\frac{x+y}{1-xy}=\large\frac{\large\frac{3}{4}+\large\frac{4}{3}}{1-\large\frac{3}{4}.\large\frac{4}{3}}=\large\frac{\large\frac{25}{12}}{1-1}=\infty\)
\( \Rightarrow tan^{-1}\large\frac{3}{4}+tan^{-1}\large\frac{4}{3}=tan^{-1}\infty\)
\( = -\pi+tan^{-1} \infty = -\pi+\large\frac{\pi}{2}\)
\( = \large\frac{-\pi}{2}\)

 

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