Question

If a = cos θ + i sin θ, then \[\frac{1 + a}{1 - a} =\]

Options

• \[\cot\frac{\theta}{2}\]

• cot θ

• \[i \cot\frac{\theta}{2}\]

• \[i \tan\frac{\theta}{2}\]

Answer 1

\[i \cot\frac{\theta}{2}\]

\[a = \cos\theta + i\sin\theta \left( \text { given } \right)\]

\[ \Rightarrow \frac{1 + a}{1 - a} = \frac{1 + \cos\theta + i\sin\theta}{1 - \cos\theta - i\sin\theta}\]

\[ \Rightarrow \frac{1 + a}{1 - a} = \frac{1 + \cos\theta + i\sin\theta}{1 - \cos\theta - i\sin\theta} \times \frac{1 - \cos\theta + i\sin\theta}{1 - \cos\theta + i\sin\theta}\]

\[\Rightarrow \frac{1 + a}{1 - a}=\frac{\left( 1 + i\sin\theta \right)^2 - \cos^2 \theta}{\left( 1 - \cos\theta \right)^2 - \left( i\sin\theta \right)^2}\]

\[\Rightarrow \frac{1 + a}{1 - a}=\frac{1 - \sin^2 \theta + 2i\sin\theta - \cos^2 \theta}{1 + \cos^2 \theta - 2\cos\theta + \sin^2 \theta}\]

\[\Rightarrow \frac{1 + a}{1 - a}=\frac{1 - \left( \sin^2 \theta + \cos^2 \theta \right) + 2i\sin\theta}{1 + \left( \sin^2 \theta + \cos^2 \theta \right) - 2\cos\theta}\]

\[\Rightarrow \frac{1 + a}{1 - a}=\frac{2i\sin\theta}{2(1 - \cos\theta)}\]

\[\Rightarrow $\frac{1 + a}{1 - a} =\frac{2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}}\]

\[\Rightarrow \frac{1 + a}{1 - a}=\frac{i\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}\]

\[\Rightarrow \frac{1 + a}{1 - a}=i \cot\frac{\theta}{2}\]