Question

If $C^2+S^2=1,then\frac{1+C+iS}{1+C-iS}$ is equal to

• A. C + iS
• B. C - iS
• C. S + iC
• D. S - iC

Answer 1

The correct option is A C + iS

We have, $C^2+S^2=1$
Let $C=cos\theta ,S=sin\theta$
$\therefore \frac{1+C+iS}{1+C-iS}=\frac{1+cos\theta +isin\theta }{1+cos\theta -isin\theta }=\frac{cos\frac{\theta }{2}(cos\frac{\theta }{2}+isin\frac{\theta }{2})}{cos\frac{\theta }{2}(cos\frac{\theta }{2}-isin\frac{\theta }{2})}=\frac{(cos\frac{\theta }{2}+isin\frac{\theta }{2}{)}^2}{1}=cos\theta +isin\theta =C+iS$