Question

Find the minimum value of 9Tan^2x + 4Cot^2x.

Answer 1

take the derivative and set that equal to zero

let y = 9tan^2 x + 4cot^2 x
dy/dx = 18tanx(sec^2x) + 8cotx(-csx^2x)
= 0

18(sinx)/(cosx)(1/cos^2x) + 8(cosx/sinx)(-1/sin^2x) = 0
18sinx/cos^3x) - 8cosx/sin^3x = 0
18sinx/cos^3x) = 8cosx/sin^3x
18sin^4x = 8cos^4x
sin^4x/cos^4x = 8/18
tan^4x = 4/9
tanx = ± (4/9)^.25 = ± .8165 appr.

set calculator to radians,
x = .6847 or π-.6847 or π+.6847 or 2π-.6847

sub each of those into original to see which one gives the smaller value

Answer 2

Let z=tan^2(x), z>0
F(z)=9z+4/z,
F'(z)=9-4/z^2, F'(z)=0 if z=2/3
minF(z)=F(2/3)=12

(corresponding value of x exists)