已知sin(a+P)=1,求证tan(2a+P)+tanP=0

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已知sin(a+P)=1,求证tan(2a+P)+tanP=0已知sin(a+P)=1,求证tan(2a+P)+tanP=0已知sin(a+P)=1,求证tan(2a+P)+tanP=0sin(a+P

已知sin(a+P)=1,求证tan(2a+P)+tanP=0
已知sin(a+P)=1,求证tan(2a+P)+tanP=0

已知sin(a+P)=1,求证tan(2a+P)+tanP=0
sin(a+P)=1
cos(a+p)=0
sin(2a+2P)=2sin(a+P)cos(a+P)=0
tan(2a+2P)=0
tan(2a+P)
=tan[(2a+2P)-p]
=[tan(2a+2P)-tanP]/[1-tan(2a+2P)tanP]
=(0-tanP)/1
即:tan(2a+P)=-tanP
所以
tan(2a+P)+tanP=0