能写一下这道题的过程吗 10
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解：
tana=（1+sinb）/cosb
=[cos(b/2)+sin(b/2)]^2/{[cos(b/2)]^2-[sin(b/2)]^2}
=[cos(b/2)+sin(b/2)]/[cos(b/2)-sin(b/2)]
=[1+tan(b/2)]/[1-tan(b/2)]
=tan(π/4+b/2)
a∈(0,π/2), b∈(0,π/2),
∴π/4+b/2∈(π/4，π/2),
∴a=π/4+b/2,
∴2a-b=π/2
tana=（1+sinb）/cosb
=[cos(b/2)+sin(b/2)]^2/{[cos(b/2)]^2-[sin(b/2)]^2}
=[cos(b/2)+sin(b/2)]/[cos(b/2)-sin(b/2)]
=[1+tan(b/2)]/[1-tan(b/2)]
=tan(π/4+b/2)
a∈(0,π/2), b∈(0,π/2),
∴π/4+b/2∈(π/4，π/2),
∴a=π/4+b/2,
∴2a-b=π/2
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