已知tanx=2,tany=3,x,y均为锐角,求证:x+y=135°

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已知tanx=2,tany=3,x,y均为锐角,求证:x+y=135°
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已知tanx=2,tany=3,x,y均为锐角,求证:x+y=135°
已知tanx=2,tany=3,x,y均为锐角,求证:x+y=135°

已知tanx=2,tany=3,x,y均为锐角,求证:x+y=135°
tan(x+y)=(tanx+tany)/(1-tanxtany)=(2+3)/(1-2*3)=-1
∵0°∴x+y=135°

tan(x+y)=(tanx+tany)/(1-tanx×tany)
=5/(-5)
=-1
又因为x,y均为锐角,所以x+y=135°

tan(x+y)=(tanx+tany)/(1-tanxtany)=(2+3)/(1-2*3)=-1
∵0°∴x+y=135°

tan(x+y)=(tanx+tany)/(1-tanxtany)=(2+3)/(1-2*3)=-1
由于x、y均为锐角,所以x+y=135°