Question

find the value of log tan 12'+log tan 18' + log tan 72' + log tan 78' with explanation

Answer 1

$log(m) + log(n) = log(mn) \\ tan(90 - x) = cot \: x \\ = log(tan12.tan18.tan72.tan78) \\ = log(tan12.tan18.cot18.cot12) \\ = log(1) = 0$