已知sin(a+β)=1,求证;tan(2a+3β)-tanβ=0
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已知sin(a+β)=1,求证;tan(2a+3β)-tanβ=0
已知sin(a+β)=1,求证;tan(2a+3β)-tanβ=0
已知sin(a+β)=1,求证;tan(2a+3β)-tanβ=0
sin(a+β)=1,a+β=π/2
tan(2a+3β)-tanβ=tan[2(a+β)+β]-tanβ
=tan(π+β)-tanβ=tanβ-tanβ=0
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