When 0 is greater than X and less than π, the minimum value of the function y = (cosx) ^ 2 / cosxsinx - (SiNx) ^ 2 is

When 0 is greater than X and less than π, the minimum value of the function y = (cosx) ^ 2 / cosxsinx - (SiNx) ^ 2 is

The original problem condition should be x ∈ (0, π / 4), because if x ∈ (0, π), then TaNx ∈ R, the original function has a minimum!
f(x)
=cos²x/(cosxsinx-sin²x)
Obviously, cosx ≠ 0
The numerator and denominator are simultaneously divided by cos & # 178; X to get
f(x)
=1/(tanx-tan²x)
Let t = TaNx, ∵ x ∈ (0, π / 4), ∵ t = TaNx ∈ (0,1)
That is to find the minimum value of F (T) = 1 / (T-T & # 178;) when t ∈ (0,1)
Let g (T) = - T & # 178; + T = - (T - &# 189;) &# 178; + &# 188;
When t ∈ (0,1), G (T) is always positive, and the maximum value is obtained when t = #
The minimum value of F (T) = 1 / g (T) is obtained at t =, #,
That is, the minimum value of F (T) is 4
That is, the minimum value of F (x) in the given range is 4
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