# Cotangent Function

$\cot(\theta)= \frac{b}{a}= \frac{\cos(\theta)}{\sin(\theta)}=\frac{1}{\tan(\theta)}$

## References

### 2019

• (Wikipedia, 2019) ⇒ https://en.wikipedia.org/wiki/Trigonometric_functions#Law_of_cotangents Retrieved:2019-4-30.
• If : $\zeta = \sqrt{\frac{1}{s} (s-a)(s-b)(s-c)}\$ (the radius of the inscribed circle for the triangle)

and : $s = \frac{a+b+c}{2 }\$ (the semi-perimeter for the triangle),

then the following all form the law of cotangents[1] : $\cot{ \frac{A}{2 }} = \frac{s-a}{\zeta }\,; \qquad \cot{ \frac{B}{2 }} = \frac{s-b}{\zeta }\,; \qquad \cot{ \frac{C}{2 }} = \frac{s-c}{\zeta }$ It follows that : $\frac{\cot \dfrac{A}{2}}{s-a} = \frac{\cot \dfrac{B}{2}}{s-b} = \frac{\cot \dfrac{C}{2}}{s-c}.$ In words the theorem is: the cotangent of a half-angle equals the ratio of the semi-perimeter minus the opposite side to the said angle, to the inradius for the triangle.

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