CBSE Class 11-science Answered
4 cosec2 A + 9 sin2 A = 4/ sin2 A + 9 sin2 A = (4 + 9 sin4 A)/ sin2 A
Let F(A) = (4 + 9 sin4 A)/ sin2 A
F(A) = [(4 × 9 sin3 A. cos A) (sin2 A) - (4 + 9 sin4 A) × 2 cos A. sin A] / (sin2 A)2
This on further simplification gives
F(A) = 2cos A. (9 sin4 A - 4)/ sin3 A
To find the maximum or the minimum value, it is required to consider, 2cos A. (9 sin4 A - 4)/ sin3 A = 0
This gives cos A = 0 or sin2 A = 2/3.
The maximum value of cosec2 A is infinite. In that case sin2 A = 0.
So, maximum value of F(A) is infinite.
When cos A = 0 i.e. sin2 A = 1, then F(A) = 4 × 1 + 9 × 1 = 13
When sin2 A = 2/3, i.e. cosec2 A = 3/2, then F(A) = 4 × 3/2 + 9 × 2/3 = 12.
Among 12 and 13, the minimum value is 12.
So, the minimum value of F(A) is 12.