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.!

What is the minimum value of 16sec^2x+9cos^2x16sec^2x+9cos^2x?