The correct option is D 2π − 6
We know that tan−1(tan θ) = θ, if −π2<θ<π2.Here, θ =−6 radians which does not lie in the interval (−π2, π2).We find that (2π − 6) lies in the interval (−π2, π2) such that tan (2π − 6) = −tan 6 = tan(−6)
∴ tan−1(tan(− 6)) = tan−1(tan (2π − 6)}= 2π − 6.