已知sin(a+b)=1,求证tan(2a+b)+tanb=0,

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已知sin(a+b)=1,求证tan(2a+b)+tanb=0,

已知sin(a+b)=1,求证tan(2a+b)+tanb=0,
已知sin(a+b)=1,求证tan(2a+b)+tanb=0,

已知sin(a+b)=1,求证tan(2a+b)+tanb=0,
证明：sin(a+b)=1
→cos(a+b)=√[1-sin^2(a+b)]=0
→sin(2a+2b)=2*sin(a+b)*cos(a+b)=0
→tan(2a+2b)=sin(2a+2b)/cos(2a+2b)=0
tan(2a+b)+tanb=tan(2a+2b-b)+tanb
=[tan(2a+2b)-tanb]/[1+tan(2a+2b)tanb]+tanb
=[0-tanb]/[1+0*tanb]+tanb
=-tanb+tanb
=0

sin(a+b)=1
cos(a+b)=0
a+b=kπ+π/2
tan(2a+b)=tan(a+b+a)=tan(kπ+π/2+a)=tan(π/2+a)=-cota
tanb=tan(a+b-a)=tan(kπ+π/2-a)=tan(π/2-a)=cota
所以tan(2a+b)+tanb=-cota+cota=0

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