Let tan⁻¹√3 = x, that is tan x = √3 = tan(π/3). Therefore, x = π/3 ∈ [-π/2, π/2]
Now let sec⁻¹(-1) = y, that is sec y = -1. Therefore, y = -sec⁻¹(1) = -0. This is not in the range of the principal value of sec⁻¹. The range is [0, π] - (π/2). Instead, we find the value of y such that sec y = -1. This occurs when y = π. However, π is not in the standard range [0, π] excluding π/2. We note that sec(π) = -1 and sec(0) = 1. The principal value is defined to be in the range [0, π], excluding π/2. However, sec(π) = -1. If we restrict to the interval [0,π], excluding π/2, then sec⁻¹(-1) is not defined. Instead, let's consider the general solution, that is, y = nπ ± π. Therefore, taking n=1, we have y = π. Note that this is only valid if we consider the general solution. In this case, y = π. If we consider the principal branch of the secant function, this solution is not valid because π is not included in the standard range of the principal value of the secant function.
However, if we consider the general solution, then sec⁻¹(-1) = π. This is not the principal value. Since sec y = -1, then y = π.
Now consider tan⁻¹√3 - sec⁻¹(-1) as shown below:
tan⁻¹√3 - sec⁻¹(-1) = x - y = π/3 - π = -2π/3
Hence, tan⁻¹√3 - sec⁻¹(-1) = -2π/3