Question

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

Answer 1

Step-by-step explanation:

a)

f(x)=cosx

utC∈IR

LhLlim

x→c − RhLlim

x→c +f(x)

limf(c−h)=lim

4−10 cos(c+h)

limcos(c−h)=lim

4−10 cosccosh−sincsinhlim

4−10 cosccosh+sincsinhcosc

=cosc

∵lLh=Rhc=f(x)

⇒cosine is continuous

b)

f(x)=cosecx= sinx1

this is continuous ∀xstsinx

=0

⇒x

=nπ 1 n∈ z

so, cosecx is cont all all n∈R

stx =nπ 1 n∈ z

c)

secx= cosx1

f is cont. ∀x∈IRstcosx

=0

⇒x

=(2n+1)π/2,n∈ z

d)

f(x)=cotx=

sinx

cosx

f is cont. for aux∈IRst

sinx =0

⇒x =nπ 1 n∈ z

So,cotx is continuous at au not numbers except x=nπ

1n∈

z