Theorem: Tangent of Sum

:$\map \tan {a + b} = \dfrac {\tan a + \tan b} {1 - \tan a \tan b}$

Proof:
First we note:
{{begin-eqn}}
{{eqn | l = \cos \sum_j \theta_j + i \sin \sum_j \theta_j
      | r = \prod_j \paren {\cos \theta_j + i \sin \theta_j}
      | c = Product of Complex Numbers in Polar Form
}}
{{eqn | r = \prod_j \cos \theta_j \prod_j \paren {1 + i \tan \theta_j}
      | c = 
}}
{{eqn | r = \prod_j \cos \theta_j \paren {1 + i \tan \theta_1} \times \paren {1 + i \tan \theta_2} \times \cdots \times \paren {1 + i \tan \theta_n}
      | c = expanding the product
}}
{{eqn | r = \prod_j \cos \theta_j \leftparen {1 + i \paren {\tan \theta_1 + \tan \theta_2 + \cdots + \tan \theta_n} }
      | c = multiplying out
}}
{{eqn | o = 
      | ro= +
      | r = i^2 \paren {\tan \theta_1 \tan \theta_2 + \tan \theta_1 \tan \theta_3 + \cdots + \tan \theta_{n - 1} \tan \theta_n}
}}
{{eqn | o = 
      | ro= +
      | r = \cdots
}}
{{eqn | o = 
      | ro= +
      | r = \rightparen {i^n \tan \theta_1 \tan \theta_2 \ldots \tan \theta_n}
}}
{{eqn | r = \prod_j \cos \theta_j \sum_{k \mathop = 0}^n i^k s_k
      | c = $s_0 = 1$
}}
{{eqn | r = \prod_j \cos \theta_j \paren {1 - s_2 + s_4 - s_6 + \cdots} + \prod_j \cos \theta_j \paren {i s_1 - i s_3 + i s_5 - \cdots }
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \cos \sum_j \theta_j
      | r = \prod_j \cos \theta_j \paren {1 - s_2 + s_4 - s_6 + \cdots}
      | c = equating real parts
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \cos \sum_j \theta_j + i \sin \sum_j \theta_j
      | r = \prod_j \paren {\cos \theta_j + i \sin \theta_j}
      | c = Product of Complex Numbers in Polar Form
}}
{{eqn | ll= \leadsto
      | l = 1 + i \tan \sum_j \theta_j
      | r = \frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \prod_j \paren {1 + i \tan \theta_j}
      | c = dividing both sides by $\cos \sum_j \theta_j$
}}
{{eqn | r = \frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {1 + i \tan \theta_1} \times \paren {1 + i \tan \theta_2} \times \cdots \times \paren {1 + i \tan \theta_n}
      | c = expanding the product
}}
{{eqn | r = \prod_j \cos \theta_j \leftparen {1 + i \paren {\tan \theta_1 + \tan \theta_2 + \cdots + \tan \theta_n} }
      | c = multiplying out
}}
{{eqn | o = 
      | ro= +
      | r = i^2 \paren {\tan \theta_1 \tan \theta_2 + \tan \theta_1 \tan \theta_3 + \cdots + \tan \theta_{n - 1} \tan \theta_n}
}}
{{eqn | o = 
      | ro= +
      | r = \cdots
}}
{{eqn | o = 
      | ro= +
      | r = \rightparen {i^n \tan \theta_1 \tan \theta_2 \ldots \tan \theta_n}
}}
{{eqn | r = \prod_j \cos \theta_j \sum_{k \mathop = 0}^n i^k s_k
      | c = $s_0 = 1$
}}
{{eqn | r = \frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {i s_1 - i s_3 + i s_5 - \cdots } + \frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {1 - s_2 + s_4 - s_6 + \cdots}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {1 + i \tan \sum_j \theta_j} {\frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {1 - s_2 + s_4 - s_6 + \cdots} }
      | r = \frac {i \paren {s_1 - s_3 + s_5 - \cdots} } {1 - s_2 + s_4 - \cdots} + 1
      | c = dividing both sides by $\frac {\prod_j \cos \theta_j} {\cos \sum_j \theta_j} \paren {1 - s_2 + s_4 - s_6 + \cdots}$
}}
{{eqn | ll= \leadsto
      | l = \frac {\cos \sum_j \theta_j \paren {1 + i \tan \sum_j \theta_j} } {\prod_j \cos \theta_j \paren {1 - s_2 + s_4 - s_6 + \cdots} }
      | r = \frac {i \paren {s_1 - s_3 + s_5 - \cdots} } {1 - s_2 + s_4 - \cdots} + 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1 + i \tan \sum_j \theta_j
      | r = \frac {i \paren {s_1 - s_3 + s_5 - \cdots} } {1 - s_2 + s_4 - \cdots} + 1
      | c = canceling top and bottom per line $1$
}}
{{eqn | ll= \leadsto
      | l = \tan \sum_j \theta_j
      | r = \frac {\paren {s_1 - s_3 + s_5 - \cdots} } {1 - s_2 + s_4 - \cdots}
      | c = 
}}
{{end-eqn}}
{{qed}}