Question

If $a=\cos \theta +i\sin \theta$, then $\frac{(1+a)}{(1-a)}$ is equal to

• A. $cot\theta$
• B. $cot\left(\frac{\theta }{2}\right)$
• C. $icot\left(\frac{\theta }{2}\right)$
• D. $i\tan \left(\frac{\theta }{2}\right)$

Answer 1

The correct option is C

$icot\left(\frac{\theta }{2}\right)$

Explanation for the correct option:

Find the value of $\frac{(1+a)}{(1-a)}$:

Given, $a=\cos \theta +i\sin \theta$

Put the value of $a$ in given expression;

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

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

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

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

$=\frac{2i\sin \theta }{2(1-\cos \theta )}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{1}\right]$

$=\frac{2i\sin \left({\displaystyle \frac{\theta }{2}}\right)\cos \left(\frac{\theta }{2}\right)}{2{\sin }^2\left(\frac{\theta }{2}\right)}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(\frac{\theta }{2}\right)\mathit{c}\mathit{o}\mathit{s}\left(\frac{\theta }{2}\right)\mathbf{,}\mathbf{}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\left(\frac{\theta }{2}\right)\right]$

$=\frac{i\cos \left(\frac{\theta }{2}\right)}{\sin \left({\displaystyle \frac{\theta }{2}}\right)}$

$=\mathit{i}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\left(\frac{\theta }{2}\right)$

Hence, Option ‘C’ is Correct.