to prove
sec(2x)+tan(2x)=(1+tan(x))/(1-tan(x))sec(2x)+tan(2x)=1+tan(x)1−tan(x)
take LHS
sec(2x)+tan(2x)= 1/cos(2x) +2tan(x)/(1-tan^2(x))sec(2x)+tan(2x)=1cos(2x)+2tan(x)1−tan2(x)
=1/(cos^2(x)-sin^2(x))+2tan(x)/(1-tan^2(x))=1cos2(x)−sin2(x)+2tan(x)1−tan2(x)
Divide every term of the first fraction by cos^2(x)cos2(x)
=(1/cos^2(x))/(cos^2(x)/cos^2(x)-sin^2(x)/cos^2(x))+2tan(x)/(1-tan^2(x))1cos2(x)cos2(x)cos2(x)−sin2(x)cos2(x)+2tan(x)1−tan2(x)
=sec^2(x)/(1-tan^2(x))+2tan(x)/(1-tan^2(x))=sec2(x)1−tan2(x)+2tan(x)1−tan2(x)
=(1+tan^2(x))/(1-tan^2(x))+2tan(x)/(1-tan^2(x))=1+tan2(x)1−tan2(x)+2tan(x)1−tan2(x)
=(1+2tan(x)+tan^2(x))/(1-tan^2(x))=1+2tan(x)+tan2(x)1−tan2(x)
=(1+tan(x))^2/((1-tan(x))(1+tan(x))=(1+tan(x))2(1−tan(x))(1+tan(x))
cancelling a (1+tan(x)) (1+tan(x)) bracket
=(1+tan(x))/(1-tan(x)=1+tan(x)1−tan(x)
as required.
