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Prove that, ${\left( {\frac{{1 + \sin \theta + i\cos \theta }}{{1 + \sin \theta - i\cos \theta }}} \right)^n} =$

$\cos n\left( {\frac{\pi }{2} - \theta } \right) + i\sin n\left( {\frac{\pi }{2} - \theta } \right)$

LHS = ${\left[ {\frac{{(1 + \sin \theta  + i\cos \theta )(\sin \theta  + i\cos \theta )}}{{\{ 1 + (\sin \theta  - i\cos \theta )\} (\sin \theta  + i\cos \theta )}}} \right]^n}$

$ = {\left[ {\frac{{(1 + \sin \theta  + i\cos \theta )(\sin \theta  + i\cos \theta )}}{{(\sin \theta  + i\cos \theta ) + \{ {{\sin }^2}\theta  - {{(i\cos \theta )}^2}\} }}} \right]^n}$

$ = {\left[ {\frac{{(1 + \sin \theta  + i\cos \theta )(\sin \theta  + i\cos \theta )}}{{(\sin \theta  + i\cos \theta ) + ({{\sin }^2}\theta  + {{\cos }^2}\theta )}}} \right]^n}$

$ = {\left[ {\frac{{(1 + \sin \theta  + i\cos \theta )(\sin \theta  + i\cos \theta )}}{{(1 + \sin \theta  + i\cos \theta )}}} \right]^n}$

$ = {(\sin \theta  + i\cos \theta )^n}$

$ = {\left\{ {\cos \left( {\frac{\pi }{2} - \theta } \right) + i\sin \left( {\frac{\pi }{2} - \theta } \right)} \right\}^n}$

$ = \cos n\left( {\frac{\pi }{2} - \theta } \right) + i\sin n\left( {\frac{\pi }{2} - \theta } \right)$

= RHS