(N/A) It is known that if $g$ and $h$ are two continuous functions,then:
$i.$ $\frac{h(x)}{g(x)}, g(x) \neq 0$ is continuous.
$ii.$ $\frac{1}{g(x)}, g(x) \neq 0$ is continuous.
$iii.$ $\frac{1}{h(x)}, h(x) \neq 0$ is continuous.
First,we prove that $g(x) = \sin x$ and $h(x) = \cos x$ are continuous functions.
For $g(x) = \sin x$,let $c$ be a real number. Put $x = c + h$. As $x \to c$,$h \to 0$.
$\lim_{x \to c} \sin x = \lim_{h \to 0} \sin(c + h) = \lim_{h \to 0} [\sin c \cos h + \cos c \sin h] = \sin c(1) + \cos c(0) = \sin c = g(c)$.
Thus,$g(x) = \sin x$ is continuous for all $x \in \mathbb{R}$.
Similarly,for $h(x) = \cos x$,$\lim_{x \to c} \cos x = \lim_{h \to 0} \cos(c + h) = \lim_{h \to 0} [\cos c \cos h - \sin c \sin h] = \cos c(1) - \sin c(0) = \cos c = h(c)$.
Thus,$h(x) = \cos x$ is continuous for all $x \in \mathbb{R}$.
Now,for the other functions:
$1.$ $\cos x$ is continuous for all $x \in \mathbb{R}$.
$2.$ $\csc x = \frac{1}{\sin x}$ is continuous where $\sin x \neq 0$,i.e.,$x \neq n\pi, n \in \mathbb{Z}$.
$3.$ $\sec x = \frac{1}{\cos x}$ is continuous where $\cos x \neq 0$,i.e.,$x \neq (2n+1)\frac{\pi}{2}, n \in \mathbb{Z}$.
$4.$ $\cot x = \frac{\cos x}{\sin x}$ is continuous where $\sin x \neq 0$,i.e.,$x \neq n\pi, n \in \mathbb{Z}$.