# show that tanh 2x=cos[ilog(a+ib/a-ib)]

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You need to notice that the problem provides inconsistent informations since there is no other equation to relates x and `z=a+bi` or`z=a-bi` , hence, the only thing you can do is to consider `tanh 2x = tanh(x+x)` such that:

`tanh 2x = (2tanh x)/(1 - tanh^2 x)`

Substituting `(e^(2x) - 1)/(e^(2x) +1)` for `tanh 2x` such that:

`tanh 2x = (2(e^(2x) - 1)/(e^(2x) + 1))/(1 - ((e^(2x) - 1)^2)/((e^(2x) + 1)^2))` => `tanh 2x = (2(e^(2x) - 1)(e^(2x) +1))/((e^(2x)+1-e^(2x)+1)(e^(2x)+1+e^(2x)-1))`

`tanh2x = 2(e^(4x)-1)/(4e^(2x)) => tanh2x =(e^(4x)-1)/(2e^(2x)) `

But `tanh 2x = -i*tan i*(2x)`

`-i*tan i*(2x) =(e^(4x)-1)/(2e^(2x)) `

`-i*(2tan ix)/(1 - tan^2(ix)) = (e^(4x)-1)/(2e^(2x)) `

**Hence, the only relation you can get, under the given conditions, is `-i*(2tan ix)/(1 - tan^2(ix)) = (e^(4x)-1)/(2e^(2x)).` **