I am familiar with solving trigonometric identities using De Moivre's Theorem, where only $\sin(x)$ and $\cos(x)$ terms are involved. But could not use it to solve identities involving other ratios. For example,
(1) $\tan\left(\frac{\theta}{2}\right)\sec(x)+\tan\left(\frac{\theta}{2^2}\right)\sec\left(\frac{x}{2}\right)+...+\tan\left(\frac{\theta}{2^n}\right)\sec\left(\frac{x}{2^{n-1}}\right)$
(2) $\csc(x)+\csc(2x)+...+\csc(2^nx)$
Is there any way to simplify this kind of problems and express them in smaller terms using De Moivre's Theorem?
We have $$\sin x=\frac{ e^{ix}-e^{-ix}}{2i}$$
And therefore it's reciprocal as $$\csc x=\frac{2i}{ e^{ix}-e^{-ix}}$$ which can also be written as $$\csc x=\frac{2i.e^{ix}}{ e^{2ix}-1}$$ therefore your series becomes
$\displaystyle{\frac{2i.e^{ix}}{ e^{2ix}-1}+\frac{2i.e^{2ix}}{ e^{4ix}-1}+\frac{2i.e^{4ix}}{ e^{8ix}-1}\cdots\frac{2i.e^{2^{n}ix}}{ e^{2^{n+1}ix}-1}}$
$2i\displaystyle{(\frac{e^{ix}}{ e^{2ix}-1}+\frac{e^{2ix}}{ e^{4ix}-1}+\frac{e^{4ix}}{ e^{8ix}-1}\cdots\frac{e^{2^{n}ix}}{ e^{2^{n+1}ix}-1})}$
$2i\displaystyle{(\frac{e^{ix}+1-1}{ (e^{ix}-1)(e^{ix}+1)}+\frac{e^{2ix}+1-1}{ (e^{2ix}-1)(e^{2ix}+1)}+\frac{e^{4ix}-1+1}{ (e^{4ix}-1)(e^{4ix}+1)}\cdots\frac{e^{2^{n}ix}+1-1}{ (e^{2^{n}ix}-1)(e^{2^{n}ix}+1)})}$
$2i((\frac{1}{e^{ix}-1}-\frac{1}{e^{2ix}-1})+(\frac{1}{e^{2ix}-1}-\frac{1}{e^{4ix}-1})+(\frac{1}{e^{4ix}-1}-\frac{1}{e^{8ix}-1})\cdots (\frac{1}{e^{2^{n}ix}-1}-\frac{1}{e^{2^{n+1}ix}-1}))$
which at last simplifies to $$2i((\frac{1}{e^{ix}-1}-\frac{1}{e^{2^{n+1}ix}-1}))$$
Now you may take it forward