How to find the general and principal values of cot\(^{-1}\) x?
Let cot θ = x (- ∞ < x < ∞) then θ = cot\(^{-1}\) 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 cot θ = x then the angle α is called the principal value of cot\(^{-1}\) x.
Again, if the principal value of cot\(^{-1}\) x is α (α ≠ 0, – π/2 ≤ α ≤ π/2) then its general value = nπ + α.
Therefore, cot\(^{-1}\) x = nπ + α, where, (α ≠ 0, – π/2 ≤ α ≤ π/2) and ( - ∞ < x < ∞ ).
Examples to find the general and principal
values of arc cot x:
1. Find the General and Principal Values of cot\(^{-1}\) √3
Solution:
Let x = cot\(^{-1}\) √3
⇒ cot x = √3
⇒ cot x = tan (π/6)
⇒ x = π/6
⇒ cot\(^{-1}\) √3 = π/6
Therefore, principal value of cot\(^{-1}\) √3 is π/6 and its general value = nπ + π/6.
2. Find the General and Principal Values of cot\(^{-1}\) (- √3)
Solution:
Let x = cot\(^{-1}\) (-√3)
⇒ cot x = -√3
⇒ cot x = cot (-π/6)
⇒ x = -π/6
⇒ cot\(^{-1}\) (-√3) = -π/6
Therefore, principal value of cot\(^{-1}\) (-√3) is -π/6 and its general value = nπ - π/6.
● Inverse Trigonometric Functions
11 and 12 Grade Math
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