Answer:
Option C
Explanation:
Consider ,$\left[\frac{1+\cos \left(\frac{\pi}{12}\right)+i\sin \left(\frac{\pi}{12}\right)}{1+ \cos\left(\frac{\pi}{12}\right)-i\sin\left(\frac{\pi}{12}\right)}\right]^{72}$
=$\left[\frac{2\cos^{2} \frac{\pi}{24}+i2\sin \frac{\pi}{24}\cos \frac{\pi}{24}}{2\cos^{2} \frac{\pi}{24}-i2\sin \frac{\pi}{24}\cos \frac{\pi}{24}}\right]^{72}$
=$\left[\frac{2\cos^{} \frac{\pi}{24}\left(\cos \frac{\pi}{24}+i\sin \frac{\pi}{24}\right)}{2\cos^{} \frac{\pi}{24}\left(\cos \frac{\pi}{24}-i\sin \frac{\pi}{24}\right)}\right]^{72}$
=$\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\right]^{72}$
= $\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\times \frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)}{\left(\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}\right)}\right]^{72}$
= $\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{2}}{\left(\cos^{2} \frac{\pi}{24}-i^{2} \sin^{2} \frac{\pi}{24}\right)}\right]^{72}$
=$\left[\frac{\left(\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{2}}{\cos^{2} \frac{\pi}{24}+ \sin^{2} \frac{\pi}{24}}\right]^{72}=\left( \cos \frac{\pi}{24}+i \sin \frac{\pi}{24}\right)^{144}$
= $\cos \left(\frac{\pi}{24}\times144\right)+i\sin\left( \frac{\pi}{24}\times144\right)$
[ By Dermoivre's theorm]
=$\cos 6\pi+i \sin 6 \pi$
=1 $\left[ \because \sin n\pi=0 \forall n\in Z\right]$ and
[ $\cos n \pi= \begin{cases}1, & if n is even\\-1, & if n is odd\end{cases}$]
