What is the minimum value of #16sec^2x+9cos^2x#?

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Answer 1 · 2016-09-27T19:10:08

#"The Minimum Value of "(16sec^2x+9cos^2x)" is "24#.

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

The AGH Property : Let #A, G and H# resp. denote the

Arithmetic, Geometric and Harmonic Mean of #a,b in RR^+#. Then,

#A geG ge H#. The equality holds iff #a=b#.

So, we have #a=16sec^2x gt 0, and, b=9cos^2x gt0#.

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

Hence, #A=(a+b)/2=1/2(16sec^2x+9cos^2x)#, and,

#G=sqrt(ab)=sqrt(16sec^2x*9cos^2x)=sqrt(144*1)=12#.

#:." by, AGH Prop., "1/2(16sec^2x+9cos^2x)ge12#, or,

#16sec^2x+9cos^2x ge 24#. Therefore,

#(16sec^2x+9cos^2x)_(min)=24#

Enjoy Maths.!