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Prove that log `[tan(pi/4+(ix)/2)]=i.tan^-1(sinhx)`

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#### Solution

L.H.S =` log[tan(pi/4+(ix)/2)]`

=`log [(1+tan(ix/2))/(1-tan(ix/2))]`

=`log [1+tan ((ix)/2)]- log [1-tan ((ix)/2)]`

= `log [(1+i.tanh) x/2]-log[(1-itanh )x/2]`

We have ,

`log (a+ib)=1/2log(a^2+b^2)+i tan^-1(b/a)`

∴ = `1/2 log (1+tanh^2 x/2)+i tan^-1 (tanh x/2)-[1/2 log (1+tanh^2 x/2)- i tan^-1(tanh x/2)]`

=`2i[tan^-1 (tanh x/2)]`

`L.H.S= 2 i.tan^-1 (tanh x/2)`

`R.H.S = i.tan ^-1 (sinhx)`

We know that `sinh^-1 x=log(x+sqrt(1+x^2))`

`tanh^-1 x=1/2[log((x+1)/(1-x))]`

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

Also `sinh^-1 (tanx)=tanh^-1 (x)`

`R.H.S= itan^-1(tanh x/2)`

`log [tan(pi/4+(ix)/2)=i.tan^-1 (sinhx)]`

Concept: Logarithmic Functions

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