given 3 tana tanb=1, show that 2 cos(a+b)=cos(a-b)

given 3 tan a tanb=1

so, tan a tanb =1/3

change them in sin and cos

now             sina sinb /cos a cos b=1/3

apply componendo and dividendo

use the forluae for cos(a+b) and cos (a-b). you will get your answer.

Given  that 3 tan a.tan b=1,so write tan a and tan b as sina/cos a and sin b/cos b,then 3 sin a .sin b=cos a cos b,now take RHS or LHS and expand it and substitute 3 sina sinb in the place of cos a .cos b,you will get RHS and LHS equal,so its proved.