Find the value of tan-1 [tan (5π/6)] + cos-1 [cos (13π/6)].

1.	Find the value of tan-1 [tan (5π/6)] + cos-1 [cos (13π/6)].
Answer» 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

Answer»

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

Discussion