If a = e ^ i , then 1+ a 1- a is equal to

MCQ

If $a = e ^{ i \theta}$, then $\frac{1+ a }{1- a }$ is equal to

A
$\cot \frac{\theta}{2}$
B
$\tan \theta$
✓
$i \cot \frac{\theta}{2}$
D
$i \tan \frac{\theta}{2}$

✓

Answer

Correct option: C.

$i \cot \frac{\theta}{2}$

(C)
$\frac{1+ a }{1- a }=\frac{1+ e ^{ i \theta}}{1- e ^{ i \theta}}$
$=\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}$
$=\frac{2 \cos ^2 \frac{\theta}{2}+2 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^2 \frac{\theta}{2}-2 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$
$=\cot \frac{\theta}{2}\left[\frac{\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}}{\sin \frac{\theta}{2}-i \cos \frac{\theta}{2}}\right]$
$=\cot \frac{\theta}{2}\left[\frac{ e ^{\frac{ i \theta}{2}}}{-\cos \left(\frac{\pi}{2}+\frac{\theta}{2}\right)- i \sin \left(\frac{\pi}{2}+\frac{\theta}{2}\right)}\right]$
$=\cot \frac{\theta}{2}\left[\frac{ e ^{\frac{ i \theta}{2}}}{- e ^{ i \left(\frac{\pi}{2}+\frac{\theta}{2}\right)}}\right]$
$=-\cot \frac{\theta}{2} e ^{\frac{-1 \pi}{2}}$
$=-\cot \frac{\theta}{2}\left[\cos \frac{\pi}{2}- i \sin \frac{\pi}{2}\right]$
$=i \cot \frac{\theta}{2}$

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