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Question
If a=cosα+isinα,b=cosβ+isinβ,c=cosγ+isinγ and ab+bc+ca=−1, then cos(β−γ)+cos(γ−α)+cos(α−β)=
If a=cosα+isinα,b=cosβ+isinβ,c=cosγ+isinγ and ab+bc+ca=−1, then cos(β−γ)+cos(γ−α)+cos(α−β)=
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Solution
The correct option is B −1
We have 1a=cosα−isinα,1b=cosβ−isinβNow ab=(cosα+isinα)(cosβ−isinβ)=cos(α−β)+isin(α−β)Similarly bc=cos(β−γ)+isin(β−γ) and ca=cos(γ−α)−isin(γ−α)Putting these values in ab+bc+ca=−1, we get=[cos(β−γ)+cos(γ−α)+cos(α−β)]+i[sin(α−β)+sin(β−γ)+sin(γ−α)]=−1=−1+0iComparing real part on both sides, we getcos(β−γ)+cos(γ−α)+cos(α−β)=−1
We have 1a=cosα−isinα,1b=cosβ−isinβ
Now ab=(cosα+isinα)(cosβ−isinβ)=cos(α−β)+isin(α−β)
Similarly bc=cos(β−γ)+isin(β−γ) and ca=cos(γ−α)−isin(γ−α)
Putting these values in ab+bc+ca=−1, we get
=[cos(β−γ)+cos(γ−α)+cos(α−β)]+i[sin(α−β)+sin(β−γ)+sin(γ−α)]=−1=−1+0i
Comparing real part on both sides, we get
cos(β−γ)+cos(γ−α)+cos(α−β)=−1
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