解
tan2a
=tan[(a-b)+(a+b)]
=[tan(a-b)+tan(a+b)]/[1-tan(a-b)tan(a+b)]
=(1/3+1/2)/(1-1/6)
=5/6×6/5
=1
tan2b
=tan[(a+b)-(a-b)]
=[tan(a+b)-tan(a-b)]/[1+tan(a+b)tan(a-b)]
=[1/2-1/3)/[1+1/6)
=1/6×(6/7)
=1/7