How to find the general and principal values of sec\(^{1}\) x?
Let sec θ = x ( x  ≥ 1 i.e., x ≥ 1 or, x ≤  1) then θ = sec  1x .
Here θ has infinitely many values.
Let 0 ≤ α ≤ π, where α is (α ≠ \(\frac{π}{2}\)) nonnegative smallest numerical value of these infinite number of values and satisfies the equation sec θ = x then the angle α is called the principal value of sec\(^{1}\) x.
Again, if the principal value of sec\(^{1}\) x is α (0 < α < π) and α ≠ \(\frac{π}{2}\) then its general value = 2nπ ± α, where,  x  ≥ 1.
Therefore, sec\(^{1}\) x = 2nπ ± α, where, (0 ≤ α ≤ π),  x  ≥ 1 and α ≠ \(\frac{π}{2}\).
Examples to find the general and principal values of arc sec x:
1.Find the General and Principal Values of sec \(^{1}\) 2.
Solution:
Let x = sec\(^{1}\) 2
⇒ sec x = 2
⇒ sec x = sec \(\frac{π}{3}\)
⇒ x = \(\frac{π}{3}\)
⇒ sec\(^{1}\) 2 = \(\frac{π}{3}\)
Therefore, principal value of sec\(^{1}\) 2 is \(\frac{π}{3}\) and its general value = 2nπ ± \(\frac{π}{3}\).
2. Find the General and Principal Values of sec \(^{1}\) (2).
Solution:
Let x = sec\(^{1}\) (2)
⇒ sec x = 2
⇒ sec x = sec \(\frac{π}{3}\)
⇒ sec x = sec (π  \(\frac{π}{3}\))
⇒ sec x = sec \(\frac{2π}{3}\)
⇒ x = \(\frac{2π}{3}\)
⇒ sec\(^{1}\) (2) = \(\frac{2π}{3}\)
Therefore, principal value of sec\(^{1}\) (2) is \(\frac{2π}{3}\) and its general value = 2nπ ± \(\frac{2π}{3}\).
11 and 12 Grade Math
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