F (x) = 1 + TaNx / 1 + (TaNx) ^ 2, X belongs to [Pai / 12, Pai / 2], find the value range of F (x)

F (x) = 1 + TaNx / 1 + (TaNx) ^ 2, X belongs to [Pai / 12, Pai / 2], find the value range of F (x)

F (x) = (1 + Tan x) / (1 + Tan & # 178; x), X belongs to [π / 12, π / 2], find the value range of F (x)
f(x)=[1+(sinx/cosx)]/sec²x=(cosx+sinx)/secx=(cosx+sinx)cosx
=(√2)sin(x+π/4)cosx=(√2)×(1/2)[sin(π/4)+sin(2x+π/4)]
=(√2/2)[(√2/2)+sin(2x+π/4)]=(1/2)+(√2/2)sin(2x+π/4)
So in the interval [π / 12, π / 2], MAXF (x) = f (π / 8) = (1 / 2) + (√ 2 / 2) sin (π / 2) = (1 / 2) (1 + √ 2);
minf(x)=f(π/2)=(1/2)+(√2/2)sin(π+π/4)=(1/2)-(√2/2)sin(π/4)=1/2-1/2=0
That is, the value range is [0, (1 / 2) (1 + √ 2)]

Why 2tanx + 1 / TaNx > = 2 √ (2tanx * 1 / TaNx)
Same as above, there's nothing more to say,

The form of the basic inequality is: if a and B are all greater than 0, a + b > = 2 √ AB (the condition for the equal sign to hold: if and only if a = b)
a^2-2ab+b^2>=0
(a+b)^2>=4ab
a+b>=2√ab

(1＋2tanx－(tanx)^2)/(1＋(tanx)^2)=?

First, split it into (1-tanx ^ 2) / (1 + TaNx ^ 2) + (2tanx / (1 + TaNx ^ 2)
Then the universal formula = cos2x + sin2x is derived