If sina.sinb-cosa.cosb= 1 show that tana+tanb=0

Answer 1

cos(a+b) = (cosa)(cosb) - (sina)(sinb)

So, we are given cos(a+b) = -1

Therefore, a + b = π + 2kπ

So, tan(a + b) = tan(π + 2kπ) = 0

Therefore, tan(a + b) = [tana + tanb] / [1 - tanatanb] = 0

So, tana + tanb = 0