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Solve for x $ \sin^{-1}\frac{2\alpha}{1+\alpha^2}+ \sin^{-1}\frac{2\beta}{1+\beta^2}=2\tan^{-1}x.$

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Toolbox:
  • \( sin2A= \large\frac{2tanA}{1=tan^2A}\)
  • \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\)
put \( \alpha = tanA \: and \: \beta = tanB\)
\(\Rightarrow\: A=tan^{-1}\alpha \; \: B=tan^{-1}\beta\)
\(\large\frac{2\alpha}{1+\alpha^2}=\large\frac{2tanA}{1+tan^2A}=sin2A\)
\(\large\frac{2\beta}{1+\beta^2}=\large\frac{2tanB}{1+tan^2B}=sin2B\)
 
Substituting in L.H.S. we get
\( sin^{-1} \large\frac{2\alpha}{1+\alpha^2}+sin^{-1}\large\frac{2\beta}{1+\beta^2}\)
\( sin^{-1} sin2A+sin^{-1}\: sin2B\)
\( =2A+2B=2(A+B)=2 \bigg[ tan^{-1}\alpha+tan^{-1}\beta \bigg] \)
 
Substituting L.H.S. in the given equation , we get
\( \Rightarrow 2(tan^{-1}\alpha + tan^{-1}\beta) =2 tan^{-1}x\)
\( \Rightarrow tan^{-1}\alpha + tan^{-1}\beta = tan^{-1}x\)
By taking \(\alpha\) in the place of x and \(\beta\) in the place of y in the above forfula of \(tan^{-1}x+tan^{-1}y\)we get
\( \Rightarrow\: tan^{-1}\alpha + tan^{-1}\beta= tan^{-1}\large\frac{\alpha+\beta}{1-\alpha\beta}=tan^{-1}x\)
or \(\Rightarrow\: x=\large\frac{\alpha+\beta}{1-\alpha\beta}\)

 

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

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