Inverse trigonometric functions have restricted domain and range, [.] and |.| denotes greatest integer function and modulus function respectively, then Inverse TrigonometricDomainRangeFunctiony=sin−1xxϵ[−1,1]yϵ[−π2,π2]y=cos−1xxϵ[−1,1]yϵ[0,π]y=tan−1xxϵRyϵ(−π2,π2)y=cot−1xxϵRyϵ(0,π)y=sec−1xxϵR−(−1,1)yϵ[0,π]−{π2}y=cosec−1xxϵR−(−1,1)yϵ[−π2,π2] The solution set for [cot−1x]2−6[cot−1x]+9≤0, is

Inverse trigonometric functions have restricted domain and range, [.]

1.	Inverse trigonometric functions have restricted domain and range, [.] and \|.\| denotes greatest integer function and modulus function respectively, then Inverse TrigonometricDomainRangeFunctiony=sin−1xxϵ[−1,1]yϵ[−π2,π2]y=cos−1xxϵ[−1,1]yϵ[0,π]y=tan−1xxϵRyϵ(−π2,π2)y=cot−1xxϵRyϵ(0,π)y=sec−1xxϵR−(−1,1)yϵ[0,π]−{π2}y=cosec−1xxϵR−(−1,1)yϵ[−π2,π2] The solution set for [cot−1x]2−6[cot−1x]+9≤0, is
Answer» Inverse trigonometric functions have restricted domain and range, [.] and \|.\| denotes greatest integer function and modulus function respectively, then $\begin{matrix} I n v e r s e T r i g o n o m e t r i c & D o m a i n & R a n g e F u n c t i o n y = s i n^{- 1} x & x ϵ [- 1, 1] & y ϵ [\frac{- π}{2}, \frac{π}{2}] y = c o s^{- 1} x & x ϵ [- 1, 1] & y ϵ [0, π] y = t a n^{- 1} x & x ϵ R & y ϵ (\frac{- π}{2}, \frac{π}{2}) y = c o t^{- 1} x & x ϵ R & y ϵ (0, π) y = s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [0, π] - {\frac{π}{2}} y = c o s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [\frac{- π}{2}, \frac{π}{2}] \end{matrix}$ The solution set for $[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0$ , is