How do you prove #sec(x) + 1 + ((1-tan^2(x)) / (sec(x)-1)) = cos(x)/(1-cos(x))#?

2 Answers
Jul 10, 2016

Do some conjugate multiplication, make use of trig identities, and simplify. See below.

Explanation:

Recall the Pythagorean Identity #sin^2x+cos^2x=1#. Divide both sides by #cos^2x#:
#(sin^2x+cos^2x)/cos^2x=1/cos^2x#
#->tan^2x+1=sec^2x#
We will be making use of this important identity.

Let's focus on this expression:
#secx+1#

Note that this is equivalent to #(secx+1)/1#. Multiply the top and bottom by #secx-1# (this technique is known as conjugate multiplication):
#(secx+1)/1*(secx-1)/(secx-1)#
#->((secx+1)(secx-1))/(secx-1)#
#->(sec^2x-1)/(secx-1)#

From #tan^2x+1=sec^2x#, we see that #tan^2x=sec^2x-1#. Therefore, we can replace the numerator with #tan^2x#:
#(tan^2x)/(secx-1)#

Our problem now reads:
#(tan^2x)/(secx-1)+(1-tan^2x) / (secx-1) = cosx/(1-cosx)#

We have a common denominator, so we can add the fractions on the left hand side:
#(tan^2x)/(secx-1)+(1-tan^2x) / (secx-1) = cosx/(1-cosx)#
#->(tan^2x+1-tan^2x)/(secx-1)=cosx/(1-cosx)#

The tangents cancel:
#(cancel(tan^2x)+1-cancel(tan^2x))/(secx-1)=cosx/(1-cosx)#

Leaving us with:
#1/(secx-1)=cosx/(1-cosx)#

Since #secx=1/cosx#, we can rewrite this as:
#1/(1/cosx-1)=cosx/(1-cosx)#

Adding fractions in the denominator, we see:
#1/(1/cosx-1)=cosx/(1-cosx)#
#->1/(1/cosx-(cosx)/(cosx))=cosx/(1-cosx)#

#->1/((1-cosx)/cosx)=cosx/(1-cosx)#

Using the property #1/(a/b)=b/a#, we have:
#cosx/(1-cosx)=cosx/(1-cosx)#

And that completes the proof.

Jul 11, 2016

#LHS=(secx+1)+(1-tan^2x)/(secx-1)#

#=((secx+1)(secx-1)+1-tan^2x)/(secx-1)#

#=(sec^2x-1+1-tan^2x)/(secx-1)#

#=cosx/cosx*((sec^2x-tan^2x))/((secx-1))#

#color(red)("putting",sec^2x-tan^2x=1)#

#=cosx/(cosxsecx-cosx)#

#color(red)("putting",cosxsecx=1)#

#=cosx/(1-cosx)=RHS#