a)
f(x)=cosx
utC∈IR
LhLlimx→c−RhLlimx→c+f(x)
limf(c−h)=lim4−10cos(c+h)
limcos(c−h)=lim4−10cosccosh−sincsinh
lim4−10cosccosh+sincsinhcosc
=cosc
∵lLh=Rhc=f(x)
⇒cosine is continuous
b)
f(x)=cosecx
=1sinx
this is continuous ∀xstsinx≠0
⇒x≠nπ1n∈z/
so, cosecx is cont all all n∈R
stx≠nπ1n∈z/
c)
secx=1cosx
f is cont. ∀x∈IRstcosx≠0
⇒x≠(2n+1)π/2,n∈z/
d)
f(x)=cotx=cosxsinx
f is cont. for aux∈IRst
sinx≠0
⇒x≠nπ1n∈z/
So,cotx is continuous at au not numbers except x=nπ1n∈z/