prove that ( cos 6 theta + 6 cos 4 theta + 15 cos 2 theta +10 ) / ( cos 5 theta + 5 cos 3 theta + 10 cos theta ) = 2 cos theta.


pls explain in detail and pls dont link to any other question.

Dear Student,

Please find below the solution to the asked query :

$$\begin{gathered} \mathrm{LHS} = \frac{\cos 6\mathrm{\theta } + 6 \cos 4\mathrm{\theta } + 15 \cos 2\mathrm{\theta } + 10}{\cos 5\mathrm{\theta } + 5 \cos 3\mathrm{\theta } + 10 \cos \mathrm{\theta }} \\ =\frac{\cos 6\mathrm{\theta } + \cos 4\mathrm{\theta } + 5 \cos 4\mathrm{\theta } + 5 \cos 2\mathrm{\theta } + 10 \cos 2\mathrm{\theta } + 10 \cos 0}{\cos 5\mathrm{\theta } + 5 \cos 3\mathrm{\theta } + 10 \cos \mathrm{\theta }} \\ =\frac{2 \cos \left({\displaystyle \frac{6\mathrm{\theta } + 4\mathrm{\theta }}{2}}\right) . \cos \left({\displaystyle \frac{6\mathrm{\theta }-4\mathrm{\theta }}{2}}\right) + 5\left[2 \cos \left({\displaystyle \frac{4\mathrm{\theta }+2\mathrm{\theta }}{2}}\right) . \cos \left({\displaystyle \frac{4\mathrm{\theta }-2\mathrm{\theta }}{2}}\right)\right] + 10\left[2 \cos \left({\displaystyle \frac{2\mathrm{\theta }+0}{2}}\right) . \cos \left({\displaystyle \frac{2\mathrm{\theta }-0}{2}}\right)\right]}{\cos 5\mathrm{\theta } + 5 \cos 3\mathrm{\theta } + 10 \cos \mathrm{\theta }} \\ =\frac{2 \cos 5\mathrm{\theta } . \cos \mathrm{\theta } + 5\left[2 \cos 3\mathrm{\theta } . \cos \mathrm{\theta }\right] + 10\left[2 \cos \mathrm{\theta } . \cos \mathrm{\theta }\right]}{\cos 5\mathrm{\theta } + 5 \cos 3\mathrm{\theta } + 10 \cos \mathrm{\theta }} \\ =\frac{2 \cos \mathrm{\theta }\left(\cos 5\mathrm{\theta } + 5 \cos 3\mathrm{\theta } + 10 \cos \mathrm{\theta }\right)}{\cos 5\mathrm{\theta } + 5 \cos 3\mathrm{\theta } + 10 \cos \mathrm{\theta }} \\ =2 \cos \mathrm{\theta } \\ =\mathrm{RHS} \end{gathered}$$

Hope this would clear your doubt about the topic.

If you have any more doubts,just ask here on the forum and our experts will try to help you out as soon as possible.

Regards