已知 tanX= 1/2,tan(x-b)=-2/ ,求tan(b-2x)的值

问题描述:

已知 tanX= 1/2,tan(x-b)=-2/ ,求tan(b-2x)的值

tan(b-2x)=-tan(2x-b)=-tan(x+x-b)
=-[tanx+tan(x-b)]/[1-tanxtan(x-b)]
=-(1/2-2)/(1+2/2)
=3/4已知 tanX= 1/2,tan(x-b)=-2/5 ,求tan(b-2x)的值抱歉打少了個5 = =tan(b-2x)=-tan(2x-b)=-tan(x+x-b)=-[tanx+tan(x-b)]/[1-tanxtan(x-b)]=-(1/2-2/5)/(1+2/5/2)=-1/12