Answer 1

We know that cos⁻¹ (cos x) = x if x in `[0, pi]`, which is the principal value branch of cos ⁻¹x.

Here `(7pi)/6 !in x in [0, pi]`

Now `cos^(-1) (cos  (7pi)/6)` can be written as

cos-1cos7π6 = cos-1cosπ+π6cos-1cos7π6 = cos-1- cosπ6             as, cosπ+θ = - cos θcos-1cos7π6  = cos-1- cosπ-5π6cos-1cos7π6 = cos-1-- cos 5π6   as, cosπ-θ = - cos θ

`:. cos^(-1) (cos  (7pi)/6) = cos^(-1) (cos  (5pi)/6) = (5pi)/6`

The correct answer is B.