若a=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132+1/156,且sinθ=a(θ∈[0,π/2]),则tanθ等于
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若a=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132+1/156,且sinθ=a(θ∈[0,π/2]),则tanθ等于若a=1/2+1/6
若a=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132+1/156,且sinθ=a(θ∈[0,π/2]),则tanθ等于
若a=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132+1/156,且sinθ=a(θ∈[0,π/2]),则tanθ等于
若a=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132+1/156,且sinθ=a(θ∈[0,π/2]),则tanθ等于
a=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132+1/156
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/12-1/13)
=1-1/13
=12/13
所以sinθ=12/13,又θ∈[0,π/2]),
故cosθ=5/13,从而tanθ=sinθ/cosθ=12/5
a=1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132+1/156
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/12-1/13)
=1-1/13
=12/13
所以sinθ=12/13,又θ∈[0,π/2]),
12
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