设a∈(0,π/2),b∈(0,π/2),且tana=1+sinb/cosb,则 A,3a-B
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设a∈(0,π/2),b∈(0,π/2),且tana=1+sinb/cosb,则 A,3a-B
π/2
B,2a-B=π/2
C,3a+B=π/2
D,2a+B=π/2
π/2
B,2a-B=π/2
C,3a+B=π/2
D,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,选B.
=[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,选B.
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