Reduce tan-1(cosӨ+isinӨ) to the form of a+ib. hence show that , tan-1(e^iӨ) =nπ/2+π/4-i/2* logtan(π/4-Ө/2)

      BS GREWAL

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Reduce tan-1(cosӨ+isinӨ) to the form of a+ib. hence show that, tan-1(e^iӨ) =nπ/2+π/4-i/2* log tan(π/4-Ө/2)

 

Solution :- let tan-1(cosӨ+isinӨ)=x+iy

 Then tan(x+iy)= (cosӨ+isinӨ)

  tan(x-iy)= (cosӨ-isinӨ)

now tan2x=[tan(x+iy)+ tan(x-iy)]/1- tan(x+iy) tan(x+iy)

tan2x=2cosӨ/1-(cos^2Ө+sin^2Ө)

tan2x=2cosӨ/1-1

tan2x=∞

i.e. 2x=nx+π/2

x=nx/2+π/4

tan2iy= tan(x+iy)- tan(x-iy)

= [tan(x+iy)- tan(x-iy)]/1+ tan(x+iy)*tan(x-iy)

tan2iy=2isinӨ/2

=tanh2y=sinӨ

ð  (e^2Ө-e^-2Ө)/ (e^2Ө+e^-2Ө)= sinӨ

By applying componendo and dividendo, we get

e^2Ө/ e^-2Ө=(1+ sinӨ)/ 1-sinӨ

=[(cosΘ/2+sinӨ/2)]^2/[(cosӨ/2-sinӨ/2)]^2

e^4Ө=(1+tan Ө/2)^2/(1-tan Ө/2)^2

e^2Ө=(tanπ/4+tan Ө/2)/( tanπ/4-tan Ө/2)

= tan(π/4+Ө/2)

Ө=1/2log tan(π/4+Ө/2)

e^-2Ө=(1-tan Ө/2)/(1+tan Ө/2)

=(tanπ/4-tan Ө/2)/( tanπ/4+tan Ө/2)

e^-2Ө= tan(π/-Ө/2)

-2Ө=log tan(π/-Ө/2)

 

Ө=-1/2log tan(π/4-Ө/2)

Hence , tan-1(e^iӨ) =nπ/2+π/4-i/2* log tan(π/4-Ө/2)

hence proved