The minimum value of a `tan^2 x+b cot^2x` equals the maximum value of a `sin^ 2 theta + b cos^2 theta ` where `a gt b gt 0`
when

Options

A. a=b
B. a=2b
C. a=3b
D. a=4b

Answer

D

Explanation

Let `y=atan^2x+bcot^2x` and `z=asin^2theta +bcos^2theta ` Then
`y=(sqrtatanx-sqrtbcotx)^2+2sqrt(ab)ge2sqrt(ab)`
` rArr y_(min)=2sqrt(ab)`
and , `z= asin^2theta+bcos^2theta`
`(dz)/(d theta)=(a-b)sin2theta " and " (d^2z)/ (d theta^2) = 2(a-b)cos theta`
For z to be maxium /maxium , we must have
`(dz)/(d theta)=0 rArr sin2theta=0rArr2 theta=0, pi,2pirArrtheta=0,pi//2,pi`
Clearly ,`(d^2z)/(d theta^2)lt0 " for " theta=pi/2` Hence z is maximum for `theta=pi/2`
`therefore z_(max)=asin^2pi/2+bcos^2pi/2=a`
Now,
`y_(min)=z_(max)rArr 2sqrt(ab)=arArrsqrtb=sqrtarArra=4b`