Question

Find the value of the following:

`tan^(-1) (tan  (7x)/6)`

Answer 1

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

Here `(7pi)/6 !in (-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(2pi - 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 in (-pi/2, pi/2)`

`:. tan^(-1) (tan  (7pi)/6)`

` = tan^(-1) (tan  pi/6) = pi/6`

Answer 2

Given, `tan^-1(tan (7pi)/6)`

We know that, for x ∈ `(-pi/2, pi/2)`, `cos^-1(cosx) = x`

= `tan^-1(tan  (7pi)/6)`

`= tan^-1(tan(pi + pi/6))`

= `tan^-1(tan  pi/6)`

`= pi/6`