# how the property AM>or = GM used to get minimum value of the function......e,g for what condition of a and b does minimum value of a tan^2 x + b cot^2 x equals maximum value of a sin^2 x + b cos^2x

saiomkar kandukuri
33 Points
11 years ago

by using AM>=Gm ..

(atan^2x  + bcot^2x) /2   >=    square root of ab

min value of atan^2x +bcot^2x is 2* square root ab

now consider second equation

it can be written as (a+b)/2 + (1/2)cos2x(b-a)

now max value of this function is b

so 2(square root of ab) =b if and only if (b>a)

if a>b then max value of 2nd function is a

then 2(square root of ab ) = a

these are the ttwo conditions

Akash Kumar Dutta
98 Points
11 years ago

Dear Amaan,

for min value of first func. we have..AM>or=GM
i.e      a.tan^2x  + b.cot^2x >or = 2(a.tan2x.bcot2x)^1/2    .....[ a+b/2 .or= (ab)^1/2 ]
= 2(ab)^1/2
hence min value of first function is 2(ab)^1/2

for second func.
its in the form of a.siny + b.cosy whose max value =(a^2 + b^2)^1/2
applying this we get the max. value of the func. as= (a^2 + b^2)^1/2

hence req. condition=> 2(ab)^1/2=(a^2 + b^2)^1/2
=> 4ab = a^2 + b^2
=>  a^2 + b^2 -  4ab=0  (ANS)

Regards.