Question

If a = cos θ + i sin θ, then $\frac{1+a}{1-a}=$
(a) $\cot \frac{\mathrm{\theta }}{2}$
(b) cot θ
(c) $i\cot \frac{\mathrm{\theta }}{2}$
(d) $i\tan \frac{\mathrm{\theta }}{2}$

Answer 1

(c) $icot\frac{\theta }{2}$
$$\begin{gathered} a=\cos \theta +i\sin \theta \left(\mathrm{given}\right) \\ \Rightarrow \frac{1+a}{1-a}=\frac{1+\cos \theta +i\sin \theta }{1-\cos \theta -i\sin \theta } \\ \Rightarrow \frac{1+a}{1-a}=\frac{1+\cos \theta +i\sin \theta }{1-\cos \theta -i\sin \theta }\times \frac{1-\cos \theta +i\sin \theta }{1-\cos \theta +i\sin \theta } \end{gathered}$$

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

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

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

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