tangents of sums In the identity
$$
\cos \left( \sum_i \theta_i \right) = \sum_{\text{even }n\ge0} (-1)^{n/2} \sum_{|I|=n} \prod_{i\in I} \sin\theta_i \prod_{i\not\in I}\cos\theta_i
$$
one can prove the case of finitely many values of $i$ by induction on the number of such values, and questions of convergence are easy to treat when there are infinitely many (and similarly with sines).
In the identity
$$
\tan \left( \sum_i \theta_i \right) = \frac{e_1-e_3+e_5-\cdots}{e_0 - e_2 + e_4 -\cdots}
$$
where $e_k$ is the $k$th-degree elementary symmetric polynomial in the variables $\tan\theta_i$, the finite case is similarly routine.
What is known about convergence in the infinite case?
LATER EDIT:
I derived this odd identity that I have not seen elsewhere (so attribute it to me if you mention it in a publication, unless you find it in something earlier):
$$
\csc\left( \sum_{i=1}^n \theta_i \right) = \frac{(-1)^{\lfloor(n-1)/2\rfloor}(\csc\theta_1\cdots\cdots\csc\theta_n)}{f_{(n\operatorname{mod} 2)} - f_{(n\operatorname{mod} 2)+2} + f_{(n\operatorname{mod} 2)+4} - \cdots\cdots}
$$
where


*

*$f_k$ is the $k$th-degree elemenary symmetric polynomial in the variables $\cot\theta_i$

*$\lfloor a\rfloor$ is the greatest integer $\le a$

*$(n\operatorname{mod} 2)$ is the remainder on division of $n$ by $2$


so that the $\pm$ in the numerator is
$$
\begin{cases}
+ & \text{if $n=1$ or $2$} \\
- & \text{if $n=3$ or $4$} \\
+ & \text{if $n=5$ or $6$} \\
- & \text{if $n=7$ or $8$} \\
  & \text{etc.}
\end{cases}
$$
A funny thing about this is that to get the case $n-1$ from the case $n$, you would presumably just set $\theta_n=0$, but then the cosecant and the cotangent both blow up.  So you apply L'Hopital's rule, and fully half of the terms in the denominator vanish, if viewed as terms within $f_k$.
Can anything sensible be said about $\csc\left(\sum_{i=1}^\infty \theta_i \right)$?
And (also my own)
$$
\cot\left(\sum_{i=1}^n \theta_i\right) = (-1)^{n+1} \left( \frac{f_1-f_3+f_5-\cdots}{f_0-f_2+f_4-\cdots} \right)^{(-1)^{n+1}}.
$$
so we have $\text{even}\leftrightarrow\text{odd}$ alternation between the numerator and the denominator every time $n$ is incremented by $1$.  Similar remarks about L'Hopital apply, and the same question about infinite sums can be asked.
 A: Alright I've figured out the answer to the first question I asked above.
Suppose $\theta_1+\theta_2+\theta_3+\cdots$ is an absolutely convergent infinite series.
The function $e_k$ is the $k$th-degree elementary symmetric polynomial in $\tan\theta_1, \tan\theta_2, \tan\theta_3, \ldots$,
Above I stated an identity one side of which is the cosecant of a sum.  But there is another, far more well behaved expression that we can put on the other side of that identity, and we'll need it here:
$$
\csc\left(\sum_{i=1}^\infty \theta_i\right) = \frac{\displaystyle\prod_{i=1}^\infty \sec\theta_i}{\displaystyle e_1-e_3+e_5-e_7+\cdots}.
$$
To prove this, first do the case where only finitely many $\theta$ are non-zero, by induction on the number of such $\theta$, then it's pretty easy to think about convergence.  In the same way, we get
$$
\sec\left(\sum_{i=1}^\infty \theta_i\right) = \frac{\displaystyle\prod_{i=1}^\infty \sec\theta_i}{\displaystyle e_0-e_2+e_4-e_6+\cdots}.
$$
(Just even indices rather than odd.)
So
$$
e_1-e_3+e_5-e_7+\cdots = \frac{\displaystyle\prod_{i=1}^\infty \sec\theta_i}{\displaystyle\csc\left(\sum_{i=1}^\infty \theta_i\right)}
$$
So the expression on the left actually converges if the one on the right does.  What this means is that we need to think about
$$
\lim_{n\to\infty} \frac{\displaystyle\prod_{i=1}^n \sec\theta_i}{\displaystyle\csc\left(\sum_{i=1}^n \theta_i\right)}.
$$
The hardest part of this is the numerator and that's not so hard.  We have
$$
1\le\sec\theta \le 1+\theta^2\text{ for $\theta$ small enough}.
$$
We can use that to show the product in the numerator converges.
In going from the case of $n$ non-zero $\theta$ to $n+1$ of them, we don't just increment the number of $e_k$ that appear in the sum, but also every $e_k$ acquires more terms.  But that is not hard to deal with.
So $e_1-e_3+e_5-e_7+\cdots$ converges!
And similarly for $e_0-e_2+e_4-e_6+\cdots$.
And from there it's not hard to show that the identity for the tangent of a sum still holds if the sum is an absolutely convergent series.
