Tangent of Sum of Series of Angles

Theorem
Let $\theta_1, \theta_2, \ldots, \theta_n$ be angles.

For all $k \in \set {1, 2, 3, \ldots}$, let $s_k$ be defined as the sum of the product of $\theta_1, \theta_2, \ldots$ taken $k$ at a time:


 * $s_k = \ds \sum_{\substack {S \mathop \in \set {1, 2, \ldots, n} \\ \size S \mathop = k} } \paren {\prod_{j \mathop \in S} \tan \theta_j}$

Then:
 * $\map \tan {\theta_1 + \theta_2 + \theta_3 + \cdots + \theta_n} = \dfrac {s_1 - s_3 + s_5 - \cdots} {1 - s_2 + s_4 - \cdots}$

Proof
From Product of Complex Numbers in Polar Form:
 * $\ds \cos \sum_j \theta_j + i \sin \sum_j \theta_j = \prod_j \paren {\cos \theta_j + i \sin \theta_j}$

we have:
 * $\ds 1 + i \tan \sum_j \theta_j = \dfrac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \prod_j \paren {1 + i \tan \theta_j}$

Hence by definition of $s_k$:


 * $\ds 1 + i \tan \sum_j \theta_j = \dfrac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \sum_k i^k s_k$

The result follows on dividing the imaginary part by the real part.