Recall the Pythagorean Identity sin^2x+cos^2x=1sin2x+cos2x=1. Divide both sides by cos^2xcos2x:
(sin^2x+cos^2x)/cos^2x=1/cos^2xsin2x+cos2xcos2x=1cos2x
->tan^2x+1=sec^2x→tan2x+1=sec2x
We will be making use of this important identity.
Let's focus on this expression:
secx+1secx+1
Note that this is equivalent to (secx+1)/1secx+11. Multiply the top and bottom by secx-1secx−1 (this technique is known as conjugate multiplication):
(secx+1)/1*(secx-1)/(secx-1)secx+11⋅secx−1secx−1
->((secx+1)(secx-1))/(secx-1)→(secx+1)(secx−1)secx−1
->(sec^2x-1)/(secx-1)→sec2x−1secx−1
From tan^2x+1=sec^2xtan2x+1=sec2x, we see that tan^2x=sec^2x-1tan2x=sec2x−1. Therefore, we can replace the numerator with tan^2xtan2x:
(tan^2x)/(secx-1)tan2xsecx−1
Our problem now reads:
(tan^2x)/(secx-1)+(1-tan^2x) / (secx-1) = cosx/(1-cosx)tan2xsecx−1+1−tan2xsecx−1=cosx1−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)tan2xsecx−1+1−tan2xsecx−1=cosx1−cosx
->(tan^2x+1-tan^2x)/(secx-1)=cosx/(1-cosx)→tan2x+1−tan2xsecx−1=cosx1−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.
