A.Math Trigonometry Prove

tan (A+B) + tan (A-B) = [2sin 2A]/[cos 2A + cos 2B]
LHS
= tan (A+B) + tan (A-B)
= [sin(A+B)/cos(A+B)] + [sin(A -B)/cos(A - B)]
= [sin(A+B)cos(A -B) + sin(A -B)cos(A+B)]/[cos(A+B)cos(A - B)]
= { (1/2){sin[(A+B)+(A-B)] + sin[(A+B)-(A-B)]} + (1/2){sin[(A-B)+(A+B)] + sin[(A-B)-(A+B)]} }/[cos(A+B)cos(A - B)]
=(1/2)[sin2A + sin2B +sin2A + sin(-2B)]/[cos(A+B)cos(A - B)]
= (1/2)[sin2A + sin2B +sin2A - sin2B]/[cos(A+B)cos(A - B)]
= (1/2)(2sin2A)/[cos(A+B)cos(A - B)]
= sin2A/[cos(A+B)cos(A - B)]

RHS
= [2sin 2A]/[cos 2A + cos 2B]
= [2sin 2A]/{2cos[(2A+2B)/2]cos[(2A-2B)/2]}
= sin2A/[cos(A+B)cos(A - B)]
= LHS
therefore, tan (A+B) + tan (A-B) = [2sin 2A]/[cos 2A + cos 2B]


P.S. use the formula
sinθcosψ = 1/2[ sin(θ+ ψ) + sin(θ- ψ)]
cosθ+ cosψ = 2cos[(θ+ ψ)/2]cos[(θ- ψ)/2]