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





11 and 12 Grade Math

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