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Q.
Secant function is bijective when its domain and range are ..A.. and $R-(-1,1)$ respectively. Here A refers to
Inverse Trigonometric Functions
Solution:
If we restrict the domain of secant function to $[0, \pi]-\left\{\frac{\pi}{2}\right\}$, then it is one-one and onto with its range as the set $R-(-1,1)$. Actually, secant function restricted to any of the intervals $[-\pi, 0]-\left\{\frac{-\pi}{2}\right\},[0, \pi]-\left\{\frac{\pi}{2}\right\},[\pi, 2 \pi]-\left\{\frac{3 \pi}{2}\right\}$ etc., is bijective and its range is $R-(-1,1)$.