Question

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

• A. $icot\frac{\theta }{2}$
• B. $i\tan \frac{\theta }{2}$
• C. $i\cos \frac{\theta }{2}$
• D. $i\cos ec\frac{\theta }{2}$

Answer 1

The correct option is A

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

Explanation for correct option.

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

Then

$\begin{array}{rcl}\frac{1+a}{1-a}& =& \frac{1+\cos \theta +i\sin \theta }{1-\cos \theta +i\sin \theta }\\ & =& \frac{\left[\right(1+\cos \theta )+i\sin \theta ]}{\left[\right(1-\cos \theta )+i\sin \theta ]}\times \frac{\left[\right(1-\cos \theta )-i\sin \theta ]}{\left[\right(1-\cos \theta )-i\sin \theta ]}\end{array}$

Simplify the above equation,

$$\begin{gathered} =\frac{\left[\right(1+\cos \theta )+i\sin \theta ]}{\left[\right(1-\cos \theta )+i\sin \theta ]}\times \frac{\left[\right(1-\cos \theta )-i\sin \theta ]}{\left[\right(1-\cos \theta )-i\sin \theta ]} \\ =\frac{(1+2i\sin \theta -{\cos }^2\theta -{\sin }^2\theta )}{(1-2\cos \theta +{\cos }^2\theta )-i^2{\sin }^2\theta )} \\ =(\frac{1+2i\sin \theta -1)}{(2-2\cos \theta )} \\ =\frac{i\sin \theta }{(1-\cos \theta )} \\ =\frac{i\times 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}} \\ =icot\frac{\theta }{2} \end{gathered}$$

Hence, correct option is$\left(A\right)$