Complex Numbers 2 Question 45
46. $\operatorname{Express} \frac{1}{(1-\cos \theta)+2 i \sin \theta}$ in the form $A+i B$.
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Solution:
- Now, $\frac{1}{(1-\cos \theta)+2 i \sin \theta}=\frac{1}{2 \sin ^{2} \frac{\theta}{2}+4 i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$
$$ \begin{aligned} & =\frac{1}{2 \sin \frac{\theta}{2} \sin \frac{\theta}{2}+2 i \cos \frac{\theta}{2}} \times \frac{\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}}{\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}} \\ & =\frac{\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \sin ^{2} \frac{\theta}{2}+4 \cos ^{2} \frac{\theta}{2}} \\ & =\frac{\sin \frac{\theta}{2}-2 i \cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2} 1+3 \cos ^{2} \frac{\theta}{2}} \\ \Rightarrow A+i B & =\frac{1}{2} 1+3 \cos ^{2} \frac{\theta}{2}-i \frac{\cot ^{\frac{\theta}{2}}}{1+3 \cos ^{2} \frac{\theta}{2}} \end{aligned} $$
