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,](/uploads/image/z/7751834-26-4.jpg?t=%E5%B7%B2%E7%9F%A5sin%28a%2Bb%29%3D1%2C%E6%B1%82%E8%AF%81tan%282a%2Bb%29%2Btanb%3D0%2C)
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
→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
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