Question

If $a=cos\theta +isin\theta$, then $\frac{1+a}{1-a}$

• A. $cot\frac{\theta }{2}$
• B. $cot\theta$
• C. $icot\frac{\theta }{2}$
• D. $itan\frac{\theta }{2}$

Answer 1

The correct option is C

$icot\frac{\theta }{2}$

$a=cos\theta +isin\theta \text{given}\Rightarrow \frac{1+a}{1-a}=\frac{1+cos\theta +i\sin \theta }{1-cos\theta -isin\theta }\times \frac{1-cos\theta +isin\theta }{1-cos\theta +isin\theta }\Rightarrow \frac{1+a}{1-a}=\frac{(1+sin\theta {)}^2-co{s}^2\theta }{(1-cos\theta {)}^2-(isin\theta {)}^2}\Rightarrow \frac{1+a}{1-a}=\frac{1-si{n}^2\theta +2isin\theta -co{s}^2\theta }{1+co{s}^2\theta -2cos\theta +si{n}^2\theta }\Rightarrow \frac{1+a}{1-a}=\frac{1-(si{n}^2\theta +co{s}^2\theta )+2isin\theta }{1+(si{n}^2\theta +co{s}^2\theta )-2cos\theta }\Rightarrow \frac{1+a}{1-a}=\frac{2isin\theta }{2(1-cos\theta )}\Rightarrow \frac{1+a}{1-a}=\frac{2isin\frac{\theta }{2}cos\frac{\theta }{2}}{2si{n}^2\frac{\theta }{2}}\Rightarrow \frac{1+a}{1-a}=\frac{icos\frac{\theta }{2}}{sin\frac{\theta }{2}}\Rightarrow \frac{1+a}{1-a}=icot\frac{\theta }{2}$