Question

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

• A. $\cot \theta$
• B. $\cot \frac{\theta }{2}$
• C. $i\cot \frac{\theta }{2}$
• D. $i\tan \frac{\theta }{2}$

Answer 1

The correct option is A $\cot \theta$

$A=\cos \theta +i\sin \theta$

$\Rightarrow \frac{1+A}{1-A}$

$=\frac{1+\cos \theta +i\sin \theta }{1-\cos \theta -i\sin \theta }$

$=\frac{(1+\cos \theta +i\sin \theta )}{(1-\cos \theta -i\sin \theta )}$ $\frac{(1-\cos \theta +i\sin \theta )}{1-\cos \theta +i\sin \theta )}$

$=\frac{1+\cos \theta +i\sin \theta -{\cos }^2\theta -\cos \theta -i\sin \theta \cos \theta +i\sin \theta +i\sin \theta \cos \theta +i^2{\sin }^2\theta }{(1-\cos \theta {)}^2-(i\sin \theta {)}^2}$

$=\frac{1+2i\sin \theta -{\cos }^2\theta -{\sin }^2\theta }{(1-2\cos \theta +{\cos }^2\theta )-i^2{\sin }^2\theta }$

$=\frac{1+2i\sin \theta -1}{2-2\cos \theta }=\frac{i\sin \theta }{1-\cos \theta }$

$\Rightarrow \frac{i\times 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{1-(1-2{\sin }^2\frac{\theta }{2})}$

$\Rightarrow \frac{ii\times 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}$

$\therefore \frac{1+A}{1-A}\Rightarrow i\cot \frac{\theta }{2}$.