f(x)=1+tanX/1+(tanx)^2,x属于[派/12,派/2],求f(x)取值范围

f(x)=1+tanX/1+(tanx)^2,x属于[派/12,派/2],求f(x)取值范围
f(x)=1+tanX/1+(tanx)^2,x属于[派/12,派/2],求f(x)取值范围

f(x)=1+tanX/1+(tanx)^2,x属于[派/12,派/2],求f(x)取值范围
f(x)=(1+tanX)/(1+tan²x),x属于[π/12,π/2],求f(x)取值范围
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)
故在区间[π/12,π/2]内,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
即值域为[0,(1/2)(1+√2)]

f(x)=1+tanX/1+(tanx)^2
=（1+tanX）x cosx^2
=(cosx)^2+sinx/cosx*(cosx)^2
=(1+cos2x)/2+sinxcosx
=1+1/2*(sin2x+cos2x)
=1+√2/2*sin(2x+π/4)
x属于[π/12,π/2]，

sin(2x+π/4)∈[-根号3/2,（根号6+根号2）/4]
f（x）∈[1-根号6/4 ，（根号3+5）/4 ]

f(x)=1+tanX/1+(tanx)^2,x属于[π/12,π/2]，求f(x)取值范围
f(x)=(cos²x+sinxcosx)/(sin2x+cos2x)
=(1/2+1/2cos2x+1/2sin2x)/(sin2x+cos2x)
=(1/2)+(1/2)/(sin2x+cos2x)
=(1/2)+1/[2√2sin(2x+π/4)]
...

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f(x)=1+tanX/1+(tanx)^2,x属于[π/12,π/2]，求f(x)取值范围
f(x)=(cos²x+sinxcosx)/(sin2x+cos2x)
=(1/2+1/2cos2x+1/2sin2x)/(sin2x+cos2x)
=(1/2)+(1/2)/(sin2x+cos2x)
=(1/2)+1/[2√2sin(2x+π/4)]
0π/4<2x+π/4<3π/4
√2/22<2√2sin(2x+π/4)≤2√2
√2/4≤1/[2√2sin(2x+π/4)]<1/2
(1/2)+√2/4≤(1/2)+1/[2√2sin(2x+π/4)]≤1
值域[(1/2)+√2/4,1)

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