Evaluate tan-1(tan 12)

Given tan^-1(tan 12)

As tan^-1(tan x) = x if x ϵ [\( \dfrac{-\pi}{2},\dfrac{\pi}{2}\)]

But here x = 12 which does not belongs to above range

We know that tan (2nπ – θ) = –tan (θ)

Tan (θ – 2nπ) = tan (θ)

Here n = 2

Tan (12 – 4π) = tan (12)

Now 12 – 4π is in the given range

tan^–1 (tan 12) = 12 – 4π.