test

sin²θ + cosec²θ is always either equal to 2 or greater than 2.

Since, sin²θ + cosec²θ = (sin⁴θ + 1) /sin²θ

Now, the lowest value of θ = 0^o

In this case (sin⁴θ + 1) /sin²θ = ∞

For θ = 90^o
(sin⁴θ + 1) /sin²θ = 2

So the values of sin²θ + cosec²θ vary from 2 to infinity.