A) \[{{x}^{3}}+\frac{1}{{{x}^{3}}}\]
B) \[{{x}^{2}}+\frac{1}{{{x}^{2}}}\]
C) \[\frac{{{x}^{2}}}{1+{{x}^{2}}}\]
D) None of these
Correct Answer: A
Solution :
| \[2\cos \theta =x+\frac{1}{x}\] |
| Cubing both sides, \[8{{\cos }^{3}}\theta ={{\left( x+\frac{1}{x} \right)}^{3}}\] |
| \[8{{\cos }^{3}}\theta ={{x}^{3}}+\frac{1}{{{x}^{3}}}+3x\cdot \frac{1}{x}\left( x+\frac{1}{x} \right)\] |
| \[8{{\cos }^{3}}\theta ={{x}^{3}}+\frac{1}{{{x}^{3}}}+3\,\,(2\cos \theta )\] [from Eq.(i)] |
| \[\Rightarrow \] \[8{{\cos }^{3}}\theta -6\cos \theta ={{x}^{3}}+\frac{1}{{{x}^{3}}}\] |
| \[\Rightarrow \] \[2\,\,[4{{\cos }^{3}}\theta -3\cos \theta ]={{x}^{3}}+\frac{1}{{{x}^{3}}}\] |
| \[\Rightarrow \] \[2\,\,[\cos 3\theta ]={{x}^{3}}+\frac{1}{{{x}^{3}}}\] |
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