Question

两角和与差的三角函数 (19 20:44:12)

Answer 1

tan(60度-A)

=(tan60-tanA)/(1+tan60tanA)

=(√3+3/4)/(√3-3/4)

=[(4√3+3)/4]/[(4√3-3)/4]

=(4√3+3)/(4√3-3)

=(4√3+3)^2/[(4√3-3)(4√3+3)]

=(57+24√3)/39

=(19+8√3)/13