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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
Dear Amaan,
for min value of first func. we have..AM>or=GMi.e a.tan^2x + b.cot^2x >or = 2(a.tan2x.bcot2x)^1/2 .....[ a+b/2 .or= (ab)^1/2 ] = 2(ab)^1/2hence 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)^1/2applying 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.
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