Question

Find the value of the following:

`tan^(-1) (tan  (7pi)/6)`

Answer 1

We know that tan^–1 (tan x) = x if `x ∈ (-pi/2, pi/2)`, which is the principal value branch of tan ^–1x.

Here, `(7pi)/6 ∉ (-pi/2, pi/2)`.

Now, `tan^(-1) (tan  (7pi)/6)` can be written as:

`tan^(-1) (tan  (7pi)/6) = tan^(-1) [tan(2pi - (5pi)/6)]`   ...[tan(2π – x) = – tan x]

= `tan^(-1) [-tan ((5pi)/6)]`

= `tan^(-1) [tan (-(5pi)/6)]`

= `tan^(-1) [tan(pi - (5pi)/6)]`

= `tan^(-1) [tan(pi/6)]`, where `pi/6 ∈ (-pi/2, pi/2)`

∴ `tan^(-1) (tan  (7pi)/6)`

= `tan^(-1) (tan  pi/6)`

= `pi/6`