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

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

Let tan θ = x (- ∞ < x < ∞) then θ = tan$$^{-1}$$ x.

Here θ has infinitely many values.

Let – $$\frac{π}{2}$$ < α < $$\frac{π}{2}$$, where α is positive or negative smallest numerical value of these infinite number of values and satisfies the equation tan θ = x then the angle α is called the principal value of tan$$^{-1}$$ x.

Again, if the principal value of tan$$^{-1}$$ x is α (– $$\frac{π}{2}$$ < α < $$\frac{π}{2}$$) then its general value = nπ + α.

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

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

1. Find the General and Principal Values of tan$$^{-1}$$ (√3).

Solution:

Let x = tan$$^{-1}$$ (√3)

⇒ tan x = √3

⇒ tan x = tan $$\frac{π}{3}$$

⇒ x = $$\frac{π}{3}$$

⇒ tan$$^{-1}$$ (√3) = $$\frac{π}{3}$$

Therefore, principal value of tan$$^{-1}$$ (√3) is $$\frac{π}{3}$$ and its general value = nπ + $$\frac{π}{3}$$.

2. Find the General and Principal Values of tan$$^{-1}$$ (- √3)

Solution:

Let x = tan$$^{-1}$$ (-√3)

⇒ tan x = -√3

⇒ tan x = tan (-$$\frac{π}{3}$$)

⇒ x = -$$\frac{π}{3}$$

⇒ cos$$^{-1}$$ (-√3) = -$$\frac{π}{3}$$

Therefore, principal value of tan$$^{-1}$$ (-√3) is -$$\frac{π}{3}$$ and its general value = nπ - $$\frac{π}{3}$$.

Inverse Trigonometric Functions

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