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Question Number 68022 by anaplak last updated on 03/Sep/19

Commented by anaplak last updated on 08/Sep/19

$$yourgivensolutionisnotrightaccordingtomyquestion..$$

Answered by mr W last updated on 03/Sep/19

$$\tan (\alpha +\beta i)=x+yi$$$$\Rightarrow \frac{\sin \left(2\alpha \right)+i\sinh \left(2\beta \right)}{\cos \left(2\alpha \right)+\cosh \left(2\beta \right)}=x+yi$$$$\Rightarrow x=\frac{\sin \left(2\alpha \right)}{\cos \left(2\alpha \right)+\cosh \left(2\beta \right)}$$$$\Rightarrow y=\frac{\sinh \left(2\beta \right)}{\cos \left(2\alpha \right)+\cosh \left(2\beta \right)}$$$$\Rightarrow x^2+y^2+2x{\cot }^{-1}2\alpha =\frac{{\sin }^2\left(2\alpha \right)+{\sinh }^2\left(2\beta \right)+2\cos \left(2\alpha \right)[\cos \left(2\alpha \right)+\cosh \left(2\beta \right)]}{{[\cos \left(2\alpha \right)+\cosh \left(2\beta \right)]}^2}$$$$\Rightarrow x^2+y^2+2x{\cot }^{-1}2\alpha =\frac{1+{\sinh }^2\left(2\beta \right)+{\cos }^2\left(2\alpha \right)+2\cos \left(2\alpha \right)\cosh \left(2\beta \right)}{{[\cos \left(2\alpha \right)+\cosh \left(2\beta \right)]}^2}$$$$\Rightarrow x^2+y^2+2x{\cot }^{-1}2\alpha =\frac{{\cosh }^2\left(2\beta \right)+{\cos }^2\left(2\alpha \right)+2\cos \left(2\alpha \right)\cosh \left(2\beta \right)}{{[\cos \left(2\alpha \right)+\cosh \left(2\beta \right)]}^2}$$$$\Rightarrow x^2+y^2+2x{\cot }^{-1}2\alpha =\frac{{[\cos \left(2\alpha \right)+\cosh \left(2\beta \right)]}^2}{{[\cos \left(2\alpha \right)+\cosh \left(2\beta \right)]}^2}$$$$\Rightarrow x^2+y^2+2x{\cot }^{-1}2\alpha =1$$

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