sinA+cosA=tanA(0

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sinA+cosA=tanA(0sinA+cosA=tanA(0sinA+cosA=tanA(0tana=sina+cosa=√2sin(a+π/4)∵0<a<π/2∴π/4<a+π/4<3π/4∴1

sinA+cosA=tanA(0
sinA+cosA=tanA(0

sinA+cosA=tanA(0
tana = sina+cosa = √2sin(a+π/4)
∵ 0<a<π/2
∴ π/4<a+π/4<3π/4
∴ 1<√2sin(a+π/4)≤√2
∴ 1<tana≤√2
∴ π/4<a≤arctg√2

答案为(TT/4,TT/3)
tanA=sinA+cosA=√2sin(A+π/4)
0π/41>sin(A+π/4)>√2/2
所以,1π/4