Find the tangent of the sum of inverse tangents

Python 2, 76 72 bytes

from fractions import*
f=lambda k:Fraction(k and(k+f(k-1))/(1-k*f(k-1)))

Use the identity:

tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A) * tan(B))

We have:

f(k) = 0                                    if k = 0
     = (k + f(k - 1)) / (1 - k * f(k - 1))  if k > 0

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Thanks to Luis Mendo, save 4 bytes.

Fold[+##/(1-##)&,0,Range@#]&

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A longer, but more interesting approach (32 bytes):

Im@#/Re@#&@Product[1+n I,{n,#}]&

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Uses the OEIS formula x(n) = (x(n-1)+n)/(1-n*x(n-1)) with x(0) = 0.