A) \[\frac{\sin \,\,n\,\,\theta }{{{\sin }^{n}}\theta }\]
B) \[\frac{\cos \,\,n\,\,\theta }{{{\cos }^{n}}\theta }\]
C) \[\frac{\sin \,\,n\,\,\theta }{{{\cos }^{n}}\theta }\]
D) \[\frac{\cos \,\,n\,\,\theta }{{{\sin }^{n}}\theta }\]
Correct Answer: A
Solution :
| \[{{u}^{2}}-2u+2=0\]\[\Rightarrow \]\[u=1\pm \,\,i\,\,LHS\] |
| \[\frac{{{[(\cot \theta -1)+(1+i)]}^{n}}-{{[(\cot \theta -1)+(1-i)]}^{n}}}{2\,i}\] |
| \[=\frac{{{(\cos \theta +i\sin \theta )}^{n}}-{{(\cos \theta -i\sin \theta )}^{n}}}{{{\sin }^{n}}\theta \,\,2\,i}=\frac{2\,i\sin n\,\theta }{{{\sin }^{n}}\theta \,2\,i}\] |
| \[=\frac{\sin n\,\theta }{{{\sin }^{n}}\theta }\] |
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