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$\large\frac{1+e^{-i\theta}}{1+e^{i\theta}}$=

\[\begin{array}{1 1}(1)\cos\theta+i\sin\theta&(2)\cos\theta-i\sin\theta\\(3)\sin\theta-i\cos\theta&(4)\sin\theta+i\cos\theta\end{array}\]

1 Answer

$\large\frac{1+e^{-i\theta}}{1+e^{i\theta}}=\large\frac{1+\Large\frac{1}{e^{i \theta}}}{1+ e^{i \theta}}$
$\qquad= \large\frac{e^{i \theta}+1}{e^{i \theta}(1+e^{i \theta})}$
$\qquad= \large\frac{1}{e^{i \theta}}$$=e^{- i \theta}$
$\qquad=\cos\theta-i\sin\theta$
Hence 2 is the correct answer.
answered May 14, 2014 by meena.p
 
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