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the expression tan[ilog{a-ib/a+ib)] reduces to : ans-2ab/a^2-b^2 how explain

the expression tan[ilog{a-ib/a+ib)] reduces to : ans-2ab/a^2-b^2 how explain

Grade:12

1 Answers

kushal satya
37 Points
9 years ago

MODULUS OF A-iB/A+iB IS 1.

THUS A-iB/A+iB CAN BE WRITTEN AS ewhere Θ is its argument.

on solving we get Θ = tan-(2ab/a2-b2)

 thus,

 

tan[ilog{a-ib/a+ib)]=tan(i*log(e))=tan(i*iΘ)=tan(-Θ)=-tanΘ=2ab/a2-b2

please approve the answer

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