A) \[n\theta \]
B) \[\frac{n\theta }{2}\]
C) \[-\frac{n\theta }{2}\]
D) \[-n\theta \]
Correct Answer: A
Solution :
\[{{\left( \frac{1+\cos \theta +i\sin \theta }{1+\cos \theta -i\sin \theta } \right)}^{n}}\] \[=\left[ \frac{2{{\cos }^{2}}\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\cos }^{2}}\frac{\theta }{2}-2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right]\] \[={{\left[ \frac{\cos \frac{\theta }{2}+i\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-i\sin \frac{\theta }{2}} \right]}^{n}}\] \[={{\left[ \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right]}^{n}}{{\left[ \cos \frac{\theta }{2}-i\sin \frac{\theta }{2} \right]}^{-n}}\] \[={{\left[ \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right]}^{n}}{{\left[ {{\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}^{-1}} \right]}^{-n}}\] \[={{\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}^{n}}{{\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}^{n}}\] \[={{\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}^{2n}}\] \[=\cos n\theta +i\sin n\theta \] On comparing with\[x+iy,\] \[x=\cos n\theta ,y=\sin n\theta \] \[\arg ={{\tan }^{-1}}\left( \frac{y}{x} \right)\] \[={{\tan }^{-1}}\left( \frac{\sin n\theta }{\cos n\theta } \right)\] \[={{\tan }^{-1}}(\tan n\theta )=n\theta \]
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