Secant and Cosecant are Cofunctions

Theorem

$\text {(1)}: \quad$

$\ds \forall x \in \R, \cos x \ne 0: \, $

$\ds \sec x$

$=$

$\ds \map \csc {90 \degrees - x}$

$\text {(2)}: \quad$

$\ds \forall x \in \R, \sin x \ne 0: \, $

$\ds \csc x$

$=$

$\ds \map \sec {90 \degrees - x}$

$\sec x = \dfrac 1 {\cos x}$

Hence in order for $\sec x$ to be defined it is necessary for $\cos x \ne 0$.

Then we have:

$\Box$

$\csc x = \dfrac 1 {\sin x}$

Hence in order for $\csc x$ to be defined it is necessary for $\sin x \ne 0$.

Then we have:

$\Box$

Hence the result by definition of cofunction.

$\blacksquare$