We know that,
tan-1 tan x = x, x ∈ (-π/2, π/2)
And, here
tan-1 tan (5π/6) ≠ 5π/6 as 5π/6 ∉ (-π/2, π/2)
Also,
cos-1 cos x = x; x ∈ [0, π]
So,
cos-1 cos (13π/6) ≠ 13π/6 as 13π/6 ∉ [0, π]
Now,
tan-1 [tan (5π/6)] + cos-1 [cos (13π/6)]
= tan-1 [tan (π – π /6)] + cos-1 [cos (2π + π/6)]
= tan-1 [ -tan π /6] + cos-1 [ -cos (7π/6)]
= – tan-1 [tan π /6] + cos-1 [cos (π/6)]
= – π /6 + π /6
= 0
