#### The principal value of sec^{-1} (2) is _________.

Let sec^{-1} (2) = y. Then, sec y = 2 = sec (π/3) We know that the range of the principal value branch of sin^{-1}is [0, Π] – {π/2} and sec(π/3) = 2. Therefore, principal value of sec^{-1} (2) is π/3.

#### The principal value of sin^{-1} (0) is ____________.

Let sin^{-1} (0) = y. Then, sin y = 0 We know that the range of the principal value branch of sin^{-1}is –π/2, π/2 and sin 0 = 0 and hence, sin y = sin 0 ⇒ y = 0 ⇒ sin^{-1} (0) = 0 Therefore, principal value of sin^{-1} (0) is 0.

#### The principal value along with range of cot^{-1}x is ___________.

The domain of the cot function (cotangent function) is the set {x: x ∈ R and x ≠ nΠ, n ∈ Z} and range is R. It means that cotangent function is not defined for integral multiples of Π. If we restrict the domain of cotangent function to (0, Π), then it is bijective with its range as R. In fact, cotangent function restricted to any of the intervals (–Π, 0), (0, Π), (Π, 2Π) etc. is bijective and its range is R. Thus, cot^{-1}can be defined as a function whose domain is R and range as any of the intervals (–Π, 0), (0, Π), (Π, 2Π), etc. These intervals give different branches of the cot^{-1}function. The function with range (0, Π) is called the principal value branch of the function cot^{-1}. We thus have cot^{-1}: R → (0, Π) Hence, the principal value along with range of cot^{-1}x is (0, Π).

#### If cosec^{-1}x = y, for a ≤ y ≤ b and y ≠ {0}, the values of a and b are _______ and ________.

Given that cosec^{-1}x = y then a ≤ y ≤ b and y ≠ {0} ---------- (1) and the principal value of cosec^{-1} (x) is –π/2, π/2 – {0} ---------- (2) By comparing equations (1) and (2), a = -π/2 b = π/2 Hence, if cosec^{-1}x = y, for a ≤ y ≤ b and y ≠ {0}, the values of a and b are –π/2 and π/2.