A) \[\cos \frac{\pi }{5}\]
B) \[\operatorname{cosec}\frac{3\pi }{2}\]
C) \[\cos \frac{\pi }{12}\]
D) \[\operatorname{cosec}\frac{\pi }{12}\]
Correct Answer: B
Solution :
\[\therefore \]\[{{(1+ri)}^{3}}=\lambda (1+i)\]\[\Rightarrow \]\[1-{{r}^{3}}i+3ri-3{{r}^{2}}=\lambda +i\lambda \] |
On comparing real and imaginary parts. We get \[1-3{{r}^{2}}=\lambda \]and \[-{{r}^{3}}+3r=\lambda \] Then \[-{{r}^{3}}-3r=1-3{{r}^{2}}\] |
\[\Rightarrow \]\[{{r}^{3}}-3{{r}^{2}}-3r+1=0\] \[\Rightarrow \]\[({{r}^{3}}+1)-3r(r+1)=0\] \[\Rightarrow \]\[(r+1)({{r}^{2}}-r+1-3r)=0\] |
\[\Rightarrow \]\[(r+1)({{r}^{2}}-4r+1)=0\therefore r=-1,2\pm \sqrt{3}\]\[\Rightarrow \]\[r=\operatorname{cosec}\frac{3\pi }{2}\] |
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