已知sinθ+cosθ=1/5,θ属于(0,π),则cotθ的值是?

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已知sinθ+cosθ=1/5,θ属于(0,π),则cotθ的值是?
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已知sinθ+cosθ=1/5,θ属于(0,π),则cotθ的值是?
已知sinθ+cosθ=1/5,θ属于(0,π),则cotθ的值是?

已知sinθ+cosθ=1/5,θ属于(0,π),则cotθ的值是?
sinθ+cosθ=1/5,
两边平方,
1+2sinθcosθ=1/25,
sin2θ=-24/25,
cos2θ=±√[1-(-24/25)^2]=±7/25,
cosθ=±√[(1+cos2θ)/2]
cosθ=±4/5,
sinθ=3/5,
或cosθ=±3/5,sinθ=4/5
∵θ∈(0,π),
∴sinθ>0,
sinθ=1/5-cosθ,
cosθ1/5,
则cosθ

sinθ+cosθ=√2sin(θ+π/4)=1/5,
sin(θ+π/4)=√2/10<√2/2=sin(3π/4),
∵θ∈(0,π),即(θ+π/4)∈(π/4,5π/4)
∴3π/4<θ+π/4<π
∴cos(θ+π/4)=√(1-2/100)=-7√2/10
∴cot(θ+π/4)=-7
又cot(θ+π/4)=(1-tan...

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sinθ+cosθ=√2sin(θ+π/4)=1/5,
sin(θ+π/4)=√2/10<√2/2=sin(3π/4),
∵θ∈(0,π),即(θ+π/4)∈(π/4,5π/4)
∴3π/4<θ+π/4<π
∴cos(θ+π/4)=√(1-2/100)=-7√2/10
∴cot(θ+π/4)=-7
又cot(θ+π/4)=(1-tanθtanπ/4)/(tanθ+tanπ/4)
可得tanθ=-4/3
则cotθ=-3/4

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