Consider the range of tan−1x.tan−1x∈[−π2,π2]That means we need to find a value v, which is between−π2 and π2, such thattanv=tan(4π5)
Solving the equation, v=4π5+nπ,n∈Z
Only when n = -1, v is inside the required range.
The solution we need is v=4π5−π=−π5.
Therefore tan−1(tan(4π5))=−π5
.