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If $\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}=0=\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}$, then $\cos (\mathrm{A}-\mathrm{B})+\cos (\mathrm{B}-\mathrm{C})+\cos (\mathrm{C}-\mathrm{A})=$
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The correct answer is:
$\frac{-3}{2}$
$\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}=0$ ...(i)
$\sin A+\sin B+\sin C=0$ ...(ii)
Squaring (i) and (ii) and add,
$\begin{aligned} & \left(\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}\right)+\left(\sin ^2 \mathrm{~B}+\cos ^2 \mathrm{~B}\right)+\left(\sin ^2 \mathrm{C}+\cos ^2 \mathrm{C}\right)+ \\ & 2(\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B})+2(\cos \mathrm{B} \cos \mathrm{C}+\sin \mathrm{B} \sin \mathrm{C})+ \\ & 2(\cos \mathrm{C} \cos \mathrm{A}+\sin \mathrm{C} \sin \mathrm{A})=0 \\ & =3+2[\cos (\mathrm{A}-\mathrm{B})+\cos (\mathrm{B}-\mathrm{C})+\cos (\mathrm{C}-\mathrm{A})]=0 \\ & \therefore \cos (\mathrm{A}-\mathrm{B})+\cos (\mathrm{B}-\mathrm{C})+\cos (\mathrm{C}-\mathrm{A})=\frac{-3}{2}\end{aligned}$
$\sin A+\sin B+\sin C=0$ ...(ii)
Squaring (i) and (ii) and add,
$\begin{aligned} & \left(\sin ^2 \mathrm{~A}+\cos ^2 \mathrm{~A}\right)+\left(\sin ^2 \mathrm{~B}+\cos ^2 \mathrm{~B}\right)+\left(\sin ^2 \mathrm{C}+\cos ^2 \mathrm{C}\right)+ \\ & 2(\cos \mathrm{A} \cos \mathrm{B}+\sin \mathrm{A} \sin \mathrm{B})+2(\cos \mathrm{B} \cos \mathrm{C}+\sin \mathrm{B} \sin \mathrm{C})+ \\ & 2(\cos \mathrm{C} \cos \mathrm{A}+\sin \mathrm{C} \sin \mathrm{A})=0 \\ & =3+2[\cos (\mathrm{A}-\mathrm{B})+\cos (\mathrm{B}-\mathrm{C})+\cos (\mathrm{C}-\mathrm{A})]=0 \\ & \therefore \cos (\mathrm{A}-\mathrm{B})+\cos (\mathrm{B}-\mathrm{C})+\cos (\mathrm{C}-\mathrm{A})=\frac{-3}{2}\end{aligned}$
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