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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