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This section of Wikipedia's List of trigonometric identities was, as far as I know, written entirely by me. For a time, it dealt only with finite sums, and said something like this: $$ \tan(\theta_1+\cdots+\theta_n) = \frac{e_1-e_3+e_5-\cdots}{e_0-e_2+e_4-\cdots}\tag{1} $$ where $e_k$ is the $k$th-degree elementary symmetric polynomial in the variables $\tan\theta_i$, $i=1,\ldots,n$, and the number of terms on the right depends on the number of terms on the left.

An obvious question was whether this works with infinite sums, and I've thought only about the absolutely convergent case. In that case, the numerator and denominator would each have infinitely many terms. But if you're thinking about $\lim\limits_{n\to\infty}$ of the expression on the left, then notice that as $n$ grows, each term $e_k$ on the left changes. Thinking about that seemed moderately icky and I left it alone until this point of view occurred to me: We have two trigonometric identities $$ e_0-e_2+e_4-\cdots =\frac{\sec\theta_1\sec\theta_2\sec\theta_3\cdots}{\sec(\theta_1+\theta_2+\theta_3+\cdots)}\tag{even} $$ $$ e_1-e_3+e_5-\cdots =\frac{\sec\theta_1\sec\theta_2\sec\theta_3\cdots}{\csc(\theta_1+\theta_2+\theta_3+\cdots)}\tag{odd} $$ (The second one has secants in the numerator and cosecants in the denominator, so it's an odd function, like the expression on the left.) If $\sum_i\theta_i$ converges absolutely, then it's not hard to show that the expressions on the right converge as the number of terms grows, therefore so do the expressions on the left.

Then finally, $$ \frac{(\mathrm{odd})}{\mathrm{(even)}} = \frac{\sec}{\csc} = \tan $$ and you've got the infinite case of $(1)$. (I actually posted all this in an earlier question and answer on stackexchange.)

SO MY QUESTION IS whether that argument is out there in the literature somewhere. (If so, I'll cite it in the Wikipedia article.)

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