tan[arcsin(-1/3)]
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tan[arcsin(-1/3)]tan[arcsin(-1/3)]tan[arcsin(-1/3)]设A=arcsin(-1/3)则sinA=-1/3cosA=(2根号2)/3tanA=sinA/c
tan[arcsin(-1/3)]
tan[arcsin(-1/3)]
tan[arcsin(-1/3)]
设A=arcsin(-1/3)
则sinA=-1/3
cosA=(2根号2)/3
tanA=sinA/cosA=-1/2根号2=-(根号2)/4
-1.414/4
tan[arcsin(-1/3)]
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