Question

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

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

Answer 1

The correct option is C

$-1$

Finding the value of :

Given $e^{i\theta }=\cos \theta +i\sin \theta$

In triangle,

$\begin{array}{rcl}A+B+C& =& 180°\\ & =& \pi \\ e^{iA}.e^{iB}.e^{iC}& =& e^{i(A+B+C)}\\ & =& e^{i\pi }\\ & =& \cos \pi +i\sin \pi \\ & =& -1\end{array}$

Hence, option (C) is the answer.