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 :
[a]| \[{{u}^{2}}-2u+2=0\Rightarrow u=1\pm i\] |
| \[\Rightarrow \frac{{{(x+\alpha )}^{2}}-{{(x+\beta )}^{n}}}{\alpha -\beta }\] |
| \[=\frac{{{[(cot\theta -1)+(1+i)]}^{n}}-{{[(cot\theta -1)+(1-i)]}^{n}}}{2i}\] \[(\therefore cot\theta -1=x)\] |
| \[=\frac{{{(cos\theta +isin\theta )}^{n}}-{{(cos\theta -isin\theta )}^{n}}}{{{\sin }^{n}}\theta 2i}\] |
| \[=\frac{2i\sin n\theta }{{{\sin }^{n}}\theta 2i}=\frac{\sin n\theta }{{{\sin }^{n}}\theta }\] |
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