tan(α+π/3)=tan(α+β-β+π/3)=[tan(α+β)-tan(β-π/3)]/[1+tan(α+β)*tan(β-π/3)]
=(3/5-1/3)/(1+3/5*1/3)=(4/15)/(4/5)=1/3
tan(α+π/12)=tan(α+π/3-π/4)=[tan(α+π/3)-tan(π/4)]/[1+tan(α+π/3)*tan(π/4)]
=(1/3-1/2)/(1+1/3*1/2)=(-1/6)/(7/6)=-1/7
上述主要是方法,计算过程的话,自己斟酌下
没草稿,直接敲进来了
看不懂的话,HI我……