Question

Find the minimum value of 16sec²Θ+9cosec²θ.

Answer 1

We will solve this Problem with the help of the AGH Property :

The AGH Property : Let A,GandH resp. denote the

Arithmetic, Geometric and Harmonic Mean of a,b∈R+. Then,

A≥G≥H. The equality holds iff a=b.

So, we have a=16sec2x>0,and,b=9cos2x>0.

{ We have to exclude the case cosx=0, since, in that case, secx would be undefined}.

Hence, A=a+b2=12(16sec2x+9cos2x), and,

G=√ab=√16sec2x⋅9cos2x=√144⋅1=12.

∴ by, AGH Prop., 12(16sec2x+9cos2x)≥12, or,

16sec2x+9cos2x≥24. Therefore,

(16sec2x+9cos2x)min=24

hope it helps

Answer 2

Answer:

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