How do you solve this (sec-1)/(sec+1) + (cos-1)/cos+1)=0

Question

How do you solve this trig proof 

Answer 1

(sec-1)/(sec+1)+(cos-1)/(cos+1)=0

sec=1/cos

([1/cos]-1)/([1/cos]+1)+(cos-1)/(cos+1)=0

([1/cos]-[cos/cos])/([1/cos]+[cos/cos])+(cos-1)/(cos+1)=0

([1-cos/cos])/([1+cos]/cos)+(cos-1)/(cos+1)=0

in the first fraction the cos denominators will cancel

(1-cos)/(1+cos)+(cos-1)/(cos+1)=0

-(cos-1)/(cos+1)+(cos-1)/(cos+1)=0 (you can basically stop here because you have additive inverses on the left side)

[-(cos-1)+(cos-1)]/(cos+1)=0

0/(cos+1)=0

0=0