Find max and min: Y = tanx + 1/tan^2 x?

Question

please help me

Answer 1

Tanx ranges from - ∞ to + ∞
Y' = sec^2x - 2sec^2x/tan^3x
Local extrema may occur when Y' = 0
sec^2x(1 - 2/tan^3x) = 0
secx cannot reach 0, so that leaves
tan^3x = 2
tanx ≈ 1.2599210498948731647672106072782
x ≈ 51.56°, Y ≈ 1.8899
which is a minimum.