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

(a) f(x) $=$ cos x. Clearly, domain of f $=$ R

Let a be an arbitrary real number, then f(a) $=$ cos a.

$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} \cos (a - h) = \mathop {\lim }\limits_{h \to 0} (\cos a\cosh + \sin a\sinh ) = \cos a$

$\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle h \to 0} \cos (a + h) = \mathop {\lim }\limits_{h \to 0} (\cos a\cosh - \sin a\sinh ) = \cos a$
therefore, $\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)$

Hence we can say that $f(x) = \cos x$ is continuous at a for all $a \in R.$

(b) $f(x) = \cos ecx$
$\Rightarrow$ $f(x) = \cfrac{1}{{\sin x}}$ and domain of $f = R - \{ n\pi \} ,n \in I.$

Also,$f(a) = \cfrac{1}{{\sin a}}$

$\mathop {\lim }\limits_{x \to {a^ + }} \cfrac{1}{{\sin x}} = \cfrac{1}{{\mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle x \to 0} \sin (a + h)}}$

$= \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle x \to 0} \cfrac{1}{{\sin a\cosh + \cos a\sinh }} = \cfrac{1}{{\sin a\cos 0 + \cos a\sin (0)}}$

$= \cfrac{1}{{\sin a(1) + \cos a(0)}} = \cfrac{1}{{\sin a + 0}} = \cfrac{1}{{\sin a}}$

$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle x \to 0} \cfrac{1}{{\sin (a - h)}}$

$= \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle x \to 0} \cfrac{1}{{\sin a\cosh - \cos a\sinh }} = \cfrac{1}{{\sin a}}$

therefore, $\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)$

Hence we can say that cosec x is continuous at a for all $a \in R - \{ n\pi \} ,n \in I.$

(c) $f(x) = \sec x \Rightarrow f(x) = \cfrac{1}{{\cos x}}$
Clearly, domain of $f = R - \left\{ {(2n + 1)\cfrac{\pi }{2},n \in I} \right\}$

Also, $f(a) = \cfrac{1}{{\cos a}}$

$\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle h \to 0} \cfrac{1}{{\cos (a + h)}}$

$= \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle h \to 0} \cfrac{1}{{\cos a\cosh - \sin a\sinh }}$

$= \cfrac{1}{{\cos a\cos 0 - \sin a\sin 0}} = \cfrac{1}{{\cos a(1) - \sin a(0)}} = \cfrac{1}{{\cos a}}$

$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} \cfrac{1}{{\cos (a - h)}}$

$= \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} \cfrac{1}{{\cos a\cosh + \sin a\sinh }} = \cfrac{1}{{\cos a}}$
therefore, $\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)$ `

Hence we can say that sec x is continuous at a for all $a \in R - \{ (2n + 1\} \cfrac{\pi }{2},n \in I$

(d) $f(x) = \cot x$
$f(x) = \cfrac{1}{{\tan x}}$ and domain of $f = R - \{ n\pi \} ,n \in I$

$\mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle h \to 0} \cfrac{1}{{\tan (a + h)}}$

$= \mathop {\lim }\limits_{\scriptstyle x \to a + h\atop\scriptstyle h \to 0} \cfrac{1}{{\cfrac{{\tan a + \tanh }}{{1 - \tan a\tanh }}}} = \cfrac{1}{{\cfrac{{\tan a + 0}}{{1 - \tan a\tan 0}}}} = \cfrac{1}{{\cfrac{{\tan a}}{{1 - 0}}}} = \cfrac{1}{{\tan a}}$

$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} \cfrac{1}{{\tan (a - h)}} = \mathop {\lim }\limits_{\scriptstyle x \to a - h\atop\scriptstyle h \to 0} \cfrac{1}{{\cfrac{{\tan a - \tanh }}{{1 + \tan a\tanh }}}} = \cfrac{1}{{\tan a}}$

therefore, $\mathop {\lim }\limits_{x \to a - } f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)$

Hence we can say that cot x is continuous at a for all $a \in R - n\pi ,n \in I.$