Cotangent is Cosecant divided by Secant

Theorem

Let $\theta$ be an angle such that $\sin \theta \ne 0$.

Then:

$\cot \theta = \dfrac {\cosec \theta} {\sec \theta}$

where $\cot$, $\cosec$ and $\sec$ mean cotangent, cosecant and secant respectively.

$\ds \cot \theta$

$=$

$\ds \dfrac {\cos \theta} {\sin \theta}$

Cotangent is Cosine divided by Sine, which holds when $\sin \theta \ne 0$

$\ds $

$=$

$\ds \dfrac {1 / \sec \theta} {1 / \cosec \theta}$

$\ds $

$=$

$\ds \dfrac {\dfrac 1 {\sec \theta} \cosec \theta \sec \theta} {\dfrac 1 {\cosec \theta} \cosec \theta \sec \theta}$

multiplying top and bottom by $\cosec \theta \sec \theta$

$\ds $

$=$

$\ds \dfrac {\cosec \theta} {\sec \theta}$

after simplification

$\blacksquare$

1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): trigonometric function
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): trigonometric function