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$$
\text { If } a=\cos \theta+i \sin \theta \text {, then find the value of } \frac{1+a}{1-a} \text {. }
$$
\text { If } a=\cos \theta+i \sin \theta \text {, then find the value of } \frac{1+a}{1-a} \text {. }
$$
Solution:
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Verified Answer
$$
\begin{aligned}
&\text { We have } a=\cos \theta+i \sin \theta\\
&\therefore \quad \frac{1+a}{1-a}=\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}\\
&=\frac{1+2 \cos ^2 \theta / 2-1+2 i \sin \theta / 2 \cdot \cos \theta / 2}{1-1+2 \sin ^2 \theta / 2-2 i \sin \theta / 2 \cdot \cos \theta / 2}\\
&=\frac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 \sin \theta / 2(\sin \theta / 2-i \cos \theta / 2)}\\
&=-\frac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 i \sin \theta / 2(\cos \theta / 2+i \sin \theta / 2)}=-\frac{1}{i} \cot \theta / 2\\
&=\frac{+i^2}{i} \cot \theta / 2=i \cot \theta / 2 \quad\left[\because \frac{-1}{i}=\frac{i^2}{i}\right]
\end{aligned}
$$
\begin{aligned}
&\text { We have } a=\cos \theta+i \sin \theta\\
&\therefore \quad \frac{1+a}{1-a}=\frac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}\\
&=\frac{1+2 \cos ^2 \theta / 2-1+2 i \sin \theta / 2 \cdot \cos \theta / 2}{1-1+2 \sin ^2 \theta / 2-2 i \sin \theta / 2 \cdot \cos \theta / 2}\\
&=\frac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 \sin \theta / 2(\sin \theta / 2-i \cos \theta / 2)}\\
&=-\frac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 i \sin \theta / 2(\cos \theta / 2+i \sin \theta / 2)}=-\frac{1}{i} \cot \theta / 2\\
&=\frac{+i^2}{i} \cot \theta / 2=i \cot \theta / 2 \quad\left[\because \frac{-1}{i}=\frac{i^2}{i}\right]
\end{aligned}
$$
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