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Solve the following $tan^{-1} \frac{1-x}{1+x} =\frac{1}{2} tan^{-1} x,(x>0)$

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Toolbox:
  • \( 2tan^{-1}y=tan^{-1}\large\frac{2y}{1-y^2}\:\:|y|<1\)
Given $tan^{-1} \large\frac{1-x}{1+x} =\large\frac{1}{2} tan^{-1} x,(x>0)$, let's rewrite the given equation as \( 2tan^{-1}\large\frac{1-x}{1+x}=tan^{-1}x\)
 
We know that \( 2tan^{-1}y=tan^{-1}\large\frac{2y}{1-y^2}\:\:|y|<1\)
 
By taking \(y=\large\frac{1-x}{1+x}\) we get
$2y = \large \frac {2(1-x)}{1+x}$
$1-y^2 = 1 - \large (\frac{1-x}{1+x})^2 = \large\frac{(1+x)^2 - (1-x)^2}{(1+x)^2} = \large\frac{1+2x+x^2-1-x^2+2x}{(1+x)^2} = \large\frac{4x}{(1+x)^2}$
\(\large \frac{2y}{1-y^2}= \large \frac {2(1-x)}{1+x} \times \frac{(1+x)^2}{4x} =\large \frac{2(1-x^2)}{4x} \)
 
\(\Rightarrow\: 2tan^{-1}\large\frac{1-x}{1+x}=tan^{-1}\large \frac{2(1-x^2)}{4x}\)
 
Using the above formula the given equation becomes \( tan^{-1} \large\frac{2(1-x^2)}{4x}=tan^{-1}x\)
\( \Rightarrow\: \large\frac{2(1-x^2)}{4x}=x\)
\( \Rightarrow 1-x^2=2x^2\)
 
\(\Rightarrow1=3x^2\)
\( \Rightarrow x=\large\frac{1}{\sqrt 3}\)

 

answered Feb 23, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 
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Toolbox:
  • \(tan(x-y)=\frac{tanx-tany}{1+tanx.tany}\)
  • \(tan\frac{\pi}{4}=1\)
Let \(x=tan\theta,\:\Rightarrow\:\theta=tan^{-1}x\)
Substituting the value of x in the given equation, we have
\(tan^{-1}\bigg(\frac{1-tan\theta}{1+tan\theta}\bigg)=\frac{1}{2}tan^{-1}tan\theta\)
We know that \(tan\frac{\pi}{4}=1\)
\(\Rightarrow\:\large\:tan^{-1}\bigg(\large\frac{tan\frac{\pi}{4}-tan\theta}{1+tan\frac{\pi}{4}.tan\theta}\bigg)=\large\frac{1}{2}.\theta\)
\(\Rightarrow\:\large\:tan^{-1}tan(\frac{\pi}{4}-\theta)=\large\:\frac{1}{2}\theta\)
\(\Rightarrow\:\large\frac{\pi}{4}-\theta=\large\frac{\theta}{2}\)
\(\Rightarrow\:\large\frac{3\theta}{2}=\large\frac{\pi}{4}\)
\(\Rightarrow\:\large\theta=\large\frac{\pi}{6}=\large\:tan^{-1}x\)
\(\Rightarrow\:\large\:x=\large\:tan\frac{\pi}{6}=\large\frac{1}{\sqrt3}\)
answered Mar 21, 2013 by rvidyagovindarajan_1
 

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