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]\[{{u}^{2}}-2u+2=0\Rightarrow u=1\pm i\] |
\[\Rightarrow \frac{{{(x+\alpha )}^{2}}-{{(x+\beta )}^{n}}}{\alpha -\beta }\] |
\[=\frac{{{[(cot\theta -1)+(1+i)]}^{n}}-{{[(cot\theta -1)+(1-i)]}^{n}}}{2i}\] \[(\therefore cot\theta -1=x)\] |
\[=\frac{{{(cos\theta +isin\theta )}^{n}}-{{(cos\theta -isin\theta )}^{n}}}{{{\sin }^{n}}\theta 2i}\] |
\[=\frac{2i\sin n\theta }{{{\sin }^{n}}\theta 2i}=\frac{\sin n\theta }{{{\sin }^{n}}\theta }\] |
You need to login to perform this action.
You will be redirected in
3 sec