( 1+ ( 12 )+i ( 12 ) 1+ ( 12 )-i ( 12 ) )^ 72 is equal to

MCQ

$\left(\frac{1+\cos \left(\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{12}\right)}{1+\cos \left(\frac{\pi}{12}\right)-i \sin \left(\frac{\pi}{12}\right)}\right)^{72}$ is equal to

A
$0$
B
$-1$
✓
1
D
$\frac{1}{2}$

✓

Answer

Correct option: C.

(C)
$\left(\frac{1+\cos \left(\frac{\pi}{12}\right)+ i \sin \left(\frac{\pi}{12}\right)}{1+\cos \left(\frac{\pi}{12}\right)- i \sin \left(\frac{\pi}{12}\right)}\right)^{72}$
$=\left[\frac{2 \cos ^2 \frac{\pi}{24}+2 i \sin \frac{\pi}{24} \cos \frac{\pi}{24}}{2 \cos ^2 \frac{\pi}{24}-2 i \sin \frac{\pi}{24} \cos \frac{\pi}{24}}\right]^{72}$
$=\left[\frac{\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}}{\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}}\right]^{72}$
$=\left[\frac{e^{\frac{i \pi}{24}}}{e^{-\frac{i \pi}{24}}}\right]^{72}$
$=\left[e^{\frac{i \pi}{12}}\right]^{72}$
$= e ^{6 \pi i }$
$=\cos 6 \pi+i \sin 6 \pi$
$=1$

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