de moivre theorem and summation

By considering $\displaystyle \sum(1+\iota \tan \theta)^k$ from k=0 to k=n-1, show that

$\displaystyle \sum \cos k\theta \sec^k \theta = \cot \theta\sin n\theta\sec^n\theta$

stuck here

please help

and i also dont know how to put limits in summation :P please tell that too

Re: de moivre theorem and summation

Start by writing $\displaystyle (1+i\tan \theta)^{k}$ as

$\displaystyle \left(1+i \frac{\sin \theta}{\cos \theta}\right)^{k}=\frac{(\cos \theta + i \sin \theta)^{k}}{\cos^{k}\theta},$

and then make use of De Moivre's theorem.

Re: de moivre theorem and summation

i already did what you told but cant go any further

can you please continue after that?

Re: de moivre theorem and summation

If you do as suggested the summation can be split into two parts, one real the other imaginary.

The summation itself can be summed easily since it is a GP with common ratio $\displaystyle (1+i\tan \theta).$

Do that and then equate reals and imaginaries across the two results.

(Actually, you only need to equate reals.)