How to find the general and principal values of ccs\(^{1}\) x?
Let csc θ = x ( x ≥ 1 i.e., x ≥ 1 or, x ≤  1) then θ = csc\(^{1}\) x .
Here θ has infinitely many values.
Let – \(\frac{π}{2}\) ≤ α ≤ \(\frac{π}{2}\), where α is nonzero(α ≠ 0) positive or negative smallest numerical value of these infinite number of values and satisfies the equation csc θ = x then the angle α is called the principal value of csc\(^{1}\) x.
Again, if the principal value of csc\(^{1}\) x is α (– \(\frac{π}{2}\) < α < \(\frac{π}{2}\)) and α ≠ 0 then its general value = nπ + ( 1) n α, where,  x  ≥ 1.
Therefore, tan\(^{1}\) x = nπ + α, where, (– \(\frac{π}{2}\) < α < \(\frac{π}{2}\)),  x  ≥ 1 and ( ∞ < x < ∞).
Examples to find the general and principal values of arc csc x:
1. Find the General and Principal Values of csc \(^{1}\) (√2).
Solution:
Let x = csc\(^{1}\) (√2)
⇒ csc x = √2
⇒ csc x = csc \(\frac{π}{4}\)
⇒ x = \(\frac{π}{4}\)
⇒ csc\(^{1}\) (√2) = \(\frac{π}{4}\)
Therefore, principal value of csc\(^{1}\) (√2) is \(\frac{π}{4}\) and its general value = nπ + ( 1)\(^{n}\) ∙ \(\frac{π}{4}\).
2. Find the General and Principal Values of csc \(^{1}\) (√2).
Solution:
Let x = csc\(^{1}\) (√2)
⇒ csc x = √2
⇒ csc x = csc (\(\frac{π}{4}\))
⇒ x = \(\frac{π}{4}\)
⇒ csc\(^{1}\) (√2) = \(\frac{π}{4}\)
Therefore, principal value of csc\(^{1}\) (√2) is \(\frac{π}{4}\) and its general value = nπ + ( 1)\(^{n}\) ∙ (\(\frac{π}{4}\)) = nπ  ( 1)\(^{n}\) ∙ \(\frac{π}{4}\).
11 and 12 Grade Math
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