A) \[(2\cos \theta +1)(2\cos 2\theta +1)(2\cos 4\theta +1)\]
B) \[(\cos \theta -1)(\cos 2\theta -1)(\cos 4\theta -1)\]
C) \[(2\cos \theta -1)(2\cos 2\theta -1)(2\cos 4\theta -1)\]
D) \[(2\cos \theta +1)(2\cos 2\theta +1)(2\cos 4\theta +1)\]
E) \[(2\cos \theta -1)(2\cos 2\theta -1)(2\cos 4\theta +1)\]
Correct Answer: C
Solution :
\[\frac{2\cos 8\theta +1}{2\cos 4\theta +1}=\frac{(2\cos 4\theta -1)(2\cos 4\theta +1)}{(2\cos \theta +1)}\] \[=\frac{(2\cos 4\theta -1)(2\cos 2\theta -1)(2\cos 2\theta +1)}{(2\cos \theta +1)}\] \[=\frac{\begin{align} & (2\cos 4\theta -1)(2\cos 2\theta -1)(2\cos \theta -1) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2\cos \theta +1) \\ \end{align}}{(2\cos \theta +1)}\] \[=(2\cos 4\theta -1)(2\cos 2\theta -1)(2\cos \theta -1)\]
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