Asked by | 10th Dec, 2009, 10:29: PM
sin2θ + cosec2θ is always either equal to 2 or greater than 2.
Since, sin2θ + cosec2θ = (sin4θ + 1) /sin2θ
Now, the lowest value of θ = 0o
In this case (sin4θ + 1) /sin2θ = ∞
For θ = 90o
(sin4θ + 1) /sin2θ = 2
So the values of sin2θ + cosec2θ vary from 2 to infinity.
Answered by | 11th Dec, 2009, 10:09: AM
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