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Solve the following \[2tan^{-1} (cos x) = tan^{-1} (2\, cosec\, x) \]

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Toolbox:
  • \( 2tan^{-1}y=tan^{-1}\large\frac{2y}{1-y^2}\:\:|y|<1\)
  • \(1-cos^2x=sin^2x\)
Given $2tan^{-1} (cos x) = tan^{-1} (2\, cosec\, x)$
 
By taking $y=cosx$, we get \(\large \frac{2y}{1-y^2}\)\(=\large \frac{2cosx}{1-cos^2x}\)\(=\large\frac{2cosx}{sin^2x}\)
\(\Rightarrow\: 2tan^{-1}cosx=tan^{-1}\large\frac{2cosx}{sin^2x}\)
 
Since L.H.S. = R.H.S., \( tan^{-1} \large \frac{2cosx}{1-cos^2x} \)\( = tan^{-1}(2cosecx)\)
 
\( \Rightarrow \large \frac{2cosx}{sin^2x}\)\(=2\: cosecx\)
\(\Rightarrow cosx=sin^2x.cosecx=sin^2x.\large \frac{1}{sinx}\)
 
\( \Rightarrow cosx\;sinx=sin^2x\)
\( \Rightarrow cosx=sinx\) \( \Rightarrow x=\large\frac{\pi}{4}\)

 

answered Feb 28, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 

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