We will determine the residues of the tangent and the cotangent at their poles, which by the parent entry are simple.

By the rule in the entry coefficients of Laurent series, in a simple pole $z = a$ of $f$ one has $$\mbox{Res}(f;\, a) \;=\; \lim_{z \to a}(z\!-\!a)f(z).$$

• We get first

  Res

  (1)

  
  
• All the poles of cotangent are $n\pi$ with $n \in \mathbb{Z}$ . Since $\pi$ is the period of cotangent, we could infer that the residues in all poles are the same as (1). We may also calculate (with the change of variable $z\!-\!n\pi = w$ ) directly $$\mbox{Res}(\cot;\,n\pi) \;=\; \lim_{z \to n\pi}(z\!-\!n\pi)\cot{z} \;=\; \lim_{w \to 0}w\cot(w\!+\!n\pi) \;=\; \lim_{w \to 0}w\cot{w} \;=\; 1.$$
• In the parent entry, the complement formula for the tangent function is derived. Using it, we can find the residues of tangent at its poles $\displaystyle\frac{\pi}{2}+n\pi$ , which are simple. For example, $$\mbox{Res}(\tan;\,\frac{\pi}{2}) \;=\; \lim_{z \to \frac{\pi}{2}}\left(z\!-\!\frac{\pi}{2}\right)\cot\left(\frac{\pi}{2}\!-\!z\right) \;=\; \lim_{w \to 0}w\cot(-w) \;=\; -\mbox{Res}(\cot;\,0) \;=\; -1.$$ Similarly as above, the residues in other poles are $-1$ .

Consequently, the residues of cotangent are equal to 1 and the residues of tangent equal to $-1$ .