Question

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

• A. $\frac{1}{2},-\frac{1}{2}\cot \frac{\theta }{2}$
• B. $-\frac{1}{2},-\cot \frac{\theta }{2}$
• C. $-\frac{1}{2},\frac{1}{2}\cot \frac{\theta }{2}$
• D. $+\frac{1}{2},\cot \frac{\theta }{2}$

Answer 1

The correct option is A $\frac{1}{2},-\frac{1}{2}\cot \frac{\theta }{2}$

$\frac{1}{1-cos\theta +isin\theta }$
$=\frac{1}{2si{n}^2\frac{\theta }{2}+i2sin\frac{\theta }{2}cos\frac{\theta }{2}}$
$=\frac{1}{2sin\frac{\theta }{2}}\left(\frac{1}{sin\frac{\theta }{2}+icos\frac{\theta }{2}}\right)$
$=\frac{1}{2sin\frac{\theta }{2}}(sin\frac{\theta }{2}-icos\frac{\theta }{2})$
$=\frac{1}{2}-i\frac{cot\frac{\theta }{2}}{2}$
$=a+ib$
Hence
$a=\frac{1}{2}$ and $b=\frac{-1}{2}.cot\left(\frac{\theta }{2}\right)$