Question

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

Answer 1

It is known that if g and h are two continuous functions, then
(i)\(\frac{h(x)}{g(x)}\), g(x)≠0 is continues
(ii)\(\frac{1}{g(x)}\),\(g(x)\)\(\neq\) 0 is continues
(iii)\(\frac{1}{h(x)}\),\(h(x)\neq0\) is continues
It has to be proved first that g(x)=sinx and h(x)=cosx are continuous functions. 
Let g(x)=sinx It is evident that g(x)=sinx 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
\(\lim_{x\rightarrow c}\)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)
=sin0 cos0+cos c sin0
=sin c+0
=sin c
∴\(\lim_{x\rightarrow c}\)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\(\rightarrow\)c, then h\(\rightarrow\)0
h(c)=cos c
\(\lim_{x\rightarrow c}\)h(x)=\(\lim_{x\rightarrow c}\)cosx
=\(\lim_{h\rightarrow 0}\)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-sinc\(\times\)0
=cos c
∴\(\lim_{x\rightarrow c}\)h(x)=h(c)
Therefore,h(x)=cosx is a continuous function.
It can be concluded that, 
cosec x=\(\frac{1}{sin\,x}\), sinx≠0 is continues
⇒cosec x,x≠nπ(n∈Z) is continues
Therefore,cosecant is continuous except at x=np, nÎZ
secx=\(\frac{1}{cos\,x}\),cos x≠0 is continuous
⇒sec x, x≠(2n+1)\(\frac{\pi}{2}\)(n∈Z) is continues
Therefore,secant is continuous except at x=(2n+1)\(\frac{\pi}{2}\)(n∈Z)
cotx=\(\frac{cos\,x}{sin\,x}\),sinx≠0 is continuous

⇒cotx, x≠nπ(n∈Z) is continues
Therefore, cotangent is continuous except at x=np,nÎZ