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If $2\tan^{-1}(\cos x)=\tan^{-1}(2cosecx)$,then show that $x=\Large {\frac{\pi}{4}}$

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  • \( 2tan^{-1}y=tan^{-1}\large\frac{2y}{1-y^2}\:\:|y|<1\)
  • \(1-cos^2x=sin^2x\)
By taking y=cosx, we get \(\large\frac{2y}{1-y^2}=\large\frac{2cosx}{1-cos^2x}=\large\frac{2cosx}{sin^2x}\)
\(\Rightarrow\: 2tan^{-1}cosx=tan^{-1}\large\frac{2cosx}{sin^2x}\)
\( tan^{-1} \bigg( \large\frac{2cosx}{1-cos^2x} \bigg) = tan^{-1}(2cosecx)\)
\( \Rightarrow \large\frac{2cosx}{sin^2x}=2\: cosecx\)
\(cosx=sin^2x.cosecx=sin^2x.\large\frac{1}{sinx}\)
\( \Rightarrow cosxsinx=sin^2x\)
 
\( \Rightarrow cosx=sinx\)
 
\( \Rightarrow x=\large\frac{\pi}{4}\)
answered Jun 12, 2014 by balaji.thirumalai
 
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