Cos3Q=√3/2,find tan6Q?

Answer 1

if cos3Q = √3/2,
sin3Q = ±1/2
so, tan3Q = ±1/√3

so, using the +1/2 value,

tan6Q = 2tan3Q/(1-tan^2 3Q)
= 2(1/√3)/(1 - 1/3)
= (2/√3)/(2/3)
= 3/√3
= √3

so, -√3 if sin3Q = -1/2