Question

Express $\frac{1}{(1-cos\theta +2isin\theta )}$ in the form $x+iy$

• A. $\left(\frac{1}{5+3cos\theta }\right)+\left(\frac{2cot\theta /2}{5+3cos\theta }\right)i$
• B. $\left(\frac{1}{5-3cos\theta }\right)+\left(\frac{-2cot\theta /2}{5-3cos\theta }\right)i$
• C. $\left(\frac{1}{5+3cos\theta }\right)+\left(\frac{-2cot\theta /2}{5+3cos\theta }\right)i$
• D. $\left(\frac{1}{5-3cos\theta }\right)+\left(\frac{2cot\theta /2}{5-3cos\theta }\right)i$

Answer 1

The correct option is C $\left(\frac{1}{5+3cos\theta }\right)+\left(\frac{-2cot\theta /2}{5+3cos\theta }\right)i$

To Express $\frac{1}{(1-cos\theta +2isin\theta )}$ in the form $x+iy$
We have , $\frac{1}{(1-cos\theta +2isin\theta )}$
$=\frac{1}{(1-cos\theta )+2isin\theta }\times \frac{(1-cos\theta )-2isin\theta }{(1-cos\theta )-2isin\theta }$
$=\frac{(1-cos\theta -2isin\theta )}{(1-cos\theta {)}^2+(2sin\theta {)}^2}$
$=\frac{(1-cos\theta -2isin\theta )}{1+co{s}^2\theta -2cos\theta +4si{n}^2\theta }$
$=\frac{(1-cos\theta )}{(1-cos\theta )(3cos\theta +5)}-\frac{2isin\theta }{(1-cos\theta )(3cos\theta +5)}$
$=\frac{1}{(3cos\theta +5)}-\frac{2isin\frac{\theta }{2}cos\frac{\theta }{2}}{2si{n}^2\frac{\theta }{2}(3cos\theta +5)}$
$=\frac{1}{(3cos\theta +5)}-\frac{2cot\frac{\theta }{2}}{(3cos\theta +5)}i$
$=\frac{1}{(5+3cos\theta )}+\frac{(-2cot\theta /2)}{(5+3cos\theta )}i$
Hence , Option C