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Question

If a=cosα+isinα,b=cosβ+isinβ,
c=cosγ+isinγ and ab+bc+ca=1, then

A
cos(αβ)+cos(βγ)+cos(γα)=0
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B
sin(αβ)+sin(βγ)+sin(γα)=0
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C
cos(αβ)+cos(βγ)+cos(γα)=1
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D
sin(αβ)+sin(βγ)+sin(γα)=1
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Solution

The correct option is C cos(αβ)+cos(βγ)+cos(γα)=1
Given,
a=cosα+isinα=eiαb=cosβ+isinβ=eiβ
c=cosγ+isinγ=eiγ

and
ab+bc+ca=1ei(αβ)+ei(βγ)+ei(γα)=1cos(αβ)+cos(βγ)+cos(γα)+i[sin(αβ)+sin(βγ)+sin(γα)]=1

Equating real and imaginary parts of both the sides, we get
cos(αβ)+cos(βγ)+cos(γα)=1
and
sin(αβ)+sin(βγ)+sin(γα)=0

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