If cos A + cos B = cos C and sin A + sin B = | vedclass

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(B) Given equations are:$1) \cos A + \cos B = \cos C$$2) \sin A + \sin B = \sin C$Let $z_1 = e^{iA}$,$z_2 = e^{iB}$,and $z_3 = e^{iC}$.Adding the two

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(B) Given equations are:$1) \cos A + \cos B = \cos C$$2) \sin A + \sin B = \sin C$Let $z_1 = e^{iA}$,$z_2 = e^{iB}$,and $z_3 = e^{iC}$.Adding the two equations: $(\cos A + \cos B) + i(\sin A + \sin B) = \cos C + i\sin C$This implies $e^{iA} + e^{iB} = e^{iC}$.Taking the conjugate of the equations:$\cos A + \cos B = \cos C$$-\sin A - \sin B = -\sin C$Adding these gives $e^{-iA} + e^{-iB} = e^{-iC}$.From $e^{iA} + e^{iB} = e^{iC}$,we have $e^{iA} + e^{iB} = e^{iC}$.Dividing the two conjugate forms:$\frac{e^{iA} + e^{iB}}{e^{-iA} + e^{-iB}} = \frac{e^{iC}}{e^{-iC}}$$\frac{e^{iA} + e^{iB}}{e^{-i(A+B)}(e^{iB} + e^{iA})} = e^{2iC}$$e^{i(A+B)} = e^{2iC}$Equating the imaginary parts:$\sin(A + B) = \sin 2C$Therefore,$\frac{\sin(A + B)}{\sin 2C} = 1$.

If $\cos A + \cos B = \cos C$ and $\sin A + \sin B = \sin C$,then the value of the expression $\frac{\sin(A + B)}{\sin 2C}$ is:

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