# If a = cos θ + i sin θ, then

Question:

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

(a) $\cot \frac{\theta}{2}$

(b) $\cot \theta$

(c) $i \cot \frac{\theta}{2}$

(d) $i \tan \frac{\theta}{2}$

Solution:

(c) $i \cot \frac{\theta}{2}$

$a=\cos \theta+i \sin \theta \quad$ (given )

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

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

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

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

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

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

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

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

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