# General and Principal Values of csc$$^{-1}$$  x

How to find the general and principal values of ccs$$^{-1}$$ x?

Let csc θ = x (| x |≥ 1 i.e., x ≥ 1 or, x ≤ - 1) then θ = csc$$^{-1}$$ x .

Here θ has infinitely many values.

Let – $$\frac{π}{2}$$ ≤ α ≤ $$\frac{π}{2}$$, where α is non-zero(α ≠ 0) positive or negative smallest numerical value of these infinite number of values and satisfies the equation csc θ = x then the angle α is called the principal value of csc$$^{-1}$$ x.

Again, if the principal value of csc$$^{-1}$$ x is α (– $$\frac{π}{2}$$ < α < $$\frac{π}{2}$$) and α ≠ 0 then its general value = nπ + (- 1) n α, where, | x | ≥ 1.

Therefore, tan$$^{-1}$$ x = nπ + α, where, (– $$\frac{π}{2}$$ < α < $$\frac{π}{2}$$), | x | ≥ 1 and (- ∞ < x < ∞).

Examples to find the general and principal values of arc csc x:

1. Find the General and Principal Values of csc $$^{-1}$$ (√2).

Solution:

Let x = csc$$^{-1}$$ (√2)

⇒ csc x = √2

⇒ csc x = csc $$\frac{π}{4}$$

⇒ x = $$\frac{π}{4}$$

⇒ csc$$^{-1}$$ (√2) = $$\frac{π}{4}$$

Therefore, principal value of csc$$^{-1}$$ (√2) is $$\frac{π}{4}$$ and its general value = nπ + (- 1)$$^{n}$$ ∙ $$\frac{π}{4}$$.

2. Find the General and Principal Values of csc $$^{-1}$$ (-√2).

Solution:

Let x = csc$$^{-1}$$ (-√2)

⇒ csc x = -√2

⇒ csc x = csc (-$$\frac{π}{4}$$)

⇒ x = -$$\frac{π}{4}$$

⇒ csc$$^{-1}$$ (-√2) = -$$\frac{π}{4}$$

Therefore, principal value of csc$$^{-1}$$ (-√2) is -$$\frac{π}{4}$$ and its general value = nπ + (- 1)$$^{n}$$ ∙ (-$$\frac{π}{4}$$) = nπ - (- 1)$$^{n}$$ ∙ $$\frac{π}{4}$$.

Inverse Trigonometric Functions