If a = cos theta + i sin theta,then frac{1 + | vedclass

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(C) Given $a = \cos 	heta + i \sin 	heta$.We need to evaluate $\frac{1 + a}{1 - a}$.$\frac{1 + a}{1 - a} = \frac{1 + \cos 	heta + i \sin 	heta}{1

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$i \cot \frac{	heta}{2}$

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(C) Given $a = \cos 	heta + i \sin 	heta$.We need to evaluate $\frac{1 + a}{1 - a}$.$\frac{1 + a}{1 - a} = \frac{1 + \cos 	heta + i \sin 	heta}{1 - (\cos 	heta + i \sin 	heta)} = \frac{(1 + \cos 	heta) + i \sin 	heta}{(1 - \cos 	heta) - i \sin 	heta}$.Using half-angle formulas: $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$,$1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$,and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$.$\frac{1 + a}{1 - a} = \frac{2 \cos^2 \frac{	heta}{2} + i (2 \sin \frac{	heta}{2} \cos \frac{	heta}{2})}{2 \sin^2 \frac{	heta}{2} - i (2 \sin \frac{	heta}{2} \cos \frac{	heta}{2})}$.Factor out $2 \cos \frac{	heta}{2}$ from the numerator and $2 \sin \frac{	heta}{2}$ from the denominator:$\frac{1 + a}{1 - a} = \frac{2 \cos \frac{	heta}{2} (\cos \frac{	heta}{2} + i \sin \frac{	heta}{2})}{2 \sin \frac{	heta}{2} (\sin \frac{	heta}{2} - i \cos \frac{	heta}{2})}$.Note that $\sin \frac{	heta}{2} - i \cos \frac{	heta}{2} = -i (\cos \frac{	heta}{2} + i \sin \frac{	heta}{2})$.$\frac{1 + a}{1 - a} = \frac{\cos \frac{	heta}{2} (\cos \frac{	heta}{2} + i \sin \frac{	heta}{2})}{\sin \frac{	heta}{2} (-i) (\cos \frac{	heta}{2} + i \sin \frac{	heta}{2})} = \frac{\cot \frac{	heta}{2}}{-i} = i \cot \frac{	heta}{2}$.

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

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