Question

If $e^{i\theta }=cos\theta +isin\theta ,$ then in $△ABC$ value of $e^{iA}.e^{iB}.e^{iC}$

• A. -i
• B. 1
• C. -1
• D. None of these

Answer 1

The correct option is C -1

$e^{iA}.e^{iB}.e^{iC}=e^{iA+iB+iC}=e^{i(A+B+C)}=e^{i\pi }$
$[\therefore A+B+C=\pi ]$
$=cos\pi +isin\pi =(-1)+i\left(0\right)=-1.$