Question

cos2a/1+cos2a=tga/tg2a

Answer 1

tga/(2tga/1-tg^2a)=(1-tg^2a)/2=(1-sin^2a/cos^2a)/2=(cos^2a-sin^2a)/2cos^2a=

=cos2a/2cos^2a=cos2a/(cos^2a+cos^2a-sin^2a+sin^2a)=

=cos2a/(1+cos2a)