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Question
If tan(θ+iφ)=tanα+isecα
Prove that
1)`e^(2varphi)=cot(varphi/2)`
2) `2theta=npi+pi/2+alpha`
Solution
tan(θ+iφ) = tanα+isecα ∴tan(θ-iφ)=tanα-isecα
∴tan2θ = tan[(θ+iφ)+(θ-iφ)
`=(tan(theta+ivarphi)+tan(theta-ivarphi))/(1-tan(theta+ivarphi)tan(theta-ivarphi))`
`=(tan(theta+isecalpha)+tan(theta-isecalpha))/(1-tan(theta+isecalpha)tan(theta-isecalpha)`
`=(2tanalpha)/(1-(tan^2alpha+sec^2alpha))=(2tanalpha)/(-2tan^2alpha)-cotalpha=tan(pi/2+alpha)`
`therefore 2theta=npi+pi/2+alpha.` (general value).
Again `tan(2ivarphi)=tan[(theta+ivarphi)-[(theta-ivarphi)]`
`=(tan(theta+isecalpha)+tan(theta-isecalpha))/(1+tan(theta+isecalpha)tan(theta-isecalpha)`
`therefore itan2varphi=(2isecalpha)/(2sec^2alpha)=icosalpha`
`therefore tanh2varphi=cosalpha`
`therefore 2varphi=tanh^(-1)cosalphacosalpha1/2log[(1+cosalpha)/(1-cosalpha)]`
`=1/2log[(2cos^2(alpha/2))/(2sin^2(alpha/2))]logcot(alpha/2)`
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