Sum this series: $$1+\cos{\theta}\sec{\theta}+\cos{2\theta}\sec^2{\theta}+\cdots+\cos{n\theta}\sec^n{\theta}$$ This problem comes from the Sixth Term Examination Paper III 1990.
I have proved it using induction, but the previous part of the question derived two identities: $$ \cos{\alpha}+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+2n\beta)=\frac{\sin(n+1)\beta\cos(\alpha+n\beta)}{\sin{\beta}} $$ And $$ \cos{\alpha}+\binom{n}{1}\cos(\alpha+2\beta)+\cdots+\binom{n}{r}\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta)=2^n \cos^n \beta \cos ( \alpha + n \beta) $$ I am stuck to establish a relation between these two identities and the sum of the series.
If you remember the complex numbers, you know that $$ \cos(x)=\frac{e^{ix}+e^{-ix}}{2} $$ We can sum the following series first by formulae of geometric sequences: $$ \sum_{k=0}^n e^{ik\theta} \sec^k{\theta}=\sum_{k=0}^n (e^{i\theta} \sec{\theta})^k=\frac{1-(e^{i\theta} \sec{\theta})^{n+1}}{1-e^{i\theta} \sec{\theta}} $$ Similarly, $$ \sum_{k=0}^n e^{-ik\theta} \sec^k{\theta}=\sum_{k=0}^n (e^{-i\theta} \sec{\theta})^k=\frac{1-(e^{-i\theta} \sec{\theta})^{n+1}}{1-e^{-i\theta} \sec{\theta}} $$ Therefore, $$ \sum_{k=0}^n \cos(k\theta)\sec^k(\theta)=\frac{1}{2}\left(\sum_{k=0}^n e^{-ik\theta}\sec^k{\theta}+\sum_{k=0}^n e^{ik\theta}\sec^k{\theta}\right)\\ =\frac{1}{2}\left(\frac{1-(e^{-i\theta} \sec{\theta})^{n+1}}{1-e^{-i\theta} \sec{\theta}}+\frac{1-(e^{i\theta} \sec{\theta})^{n+1}}{1-e^{i\theta} \sec{\theta}} \right)\\ =\frac{1}{2}\left(\frac{2-2\cos\theta\sec\theta+2\cos(n\theta)\sec^{n+2}\theta-2\cos((n+1)\theta)\sec^{n+1}\theta}{1-e^{i\theta}\sec\theta-e^{-i\theta}\sec\theta+\sec^2\theta} \right)\\ =\csc\theta\sec^n\theta\sin[(n+1)\theta] $$ I will let you to fil in the missing details.