Discuss the continuity of the cosine, cosecant, secant and cotangent functions,

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

It has to be proved first that $g(x)=\sin x$ and $h(x)=\cos x$ are continuous functions.

Let g (x) = sin x

It is evident that g (x) = sin x is defined for every real number.

Let c be a real number. Put x = c + h

If $x \rightarrow c$, then $h \rightarrow 0$

$g(c)=\sin c$

$\begin{aligned} \lim _{x \rightarrow c} g(x) &=\lim _{x \rightarrow c} \sin x \\ &=\lim _{h \rightarrow 0} \sin (c+h) \\ &=\lim _{h \rightarrow 0}[\sin c \cos h+\cos c \sin h] \\ &=\lim _{h \rightarrow 0}(\sin c \cos h)+\lim _{h \rightarrow 0}(\cos c \sin h) \\ &=\sin c \cos 0+\cos c \sin 0 \\ &=\sin c+0 \\ &=\sin c \end{aligned}$

$\therefore \lim _{x \rightarrow c} g(x)=g(c)$

Therefore, g is a continuous function.

Let h (x) = cos x

It is evident that h (x) = cos x is defined for every real number.

Let c be a real number. Put x = c + h

If x ® c, then h ® 0

h (c) = cos c

$\begin{aligned} \lim _{x \rightarrow c} h(x) &=\lim _{x \rightarrow c} \cos x \\ &=\lim _{h \rightarrow 0} \cos (c+h) \\ &=\lim _{h \rightarrow 0}[\cos c \cos h-\sin c \sin h] \\ &=\lim _{h \rightarrow 0} \cos c \cos h-\lim _{h \rightarrow 0} \sin c \sin h \\ &=\cos c \cos 0-\sin c \sin 0 \\ &=\cos c \times 1-\sin c \times 0 \\ &=\cos c \end{aligned}$

$\therefore \lim _{x \rightarrow c} h(x)=h(c)$

Therefore, h (x) = cos x is continuous function.

It can be concluded that,

$\operatorname{cosec} x=\frac{1}{\sin x}, \sin x \neq 0$ is continuous

$\Rightarrow \operatorname{cosec} x, x \neq n \pi(n \in Z)$ is continuous

Therefore, cosecant is continuous except at $x=n p, n \hat{\imath} \mathbf{Z}$

$\sec x=\frac{1}{\cos x}, \cos x \neq 0$ is continuous

$\Rightarrow \sec x, x \neq(2 n+1) \frac{\pi}{2}(n \in \mathbf{Z})$ is continuous

Therefore, secant is continuous except at $x=(2 n+1) \frac{\pi}{2}(n \in \mathbf{Z})$

$\cot x=\frac{\cos x}{\sin x}, \sin x \neq 0$ is continuous

$\Rightarrow \cot x, x \neq n \pi(n \in Z)$ is continuous

Therefore, cotangent is continuous except at x = np, n Î Z