解tan2a
=tan[(a+b）+（a-b）]
=[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]
=[1/2+1/3]/[1-1/2*1/3]
=[3+2]/(6-1)
=1
同理
tan2b
=tan[(a+b）-（a-b）]
=[tan(a+b)-tan(a-b)]/[1+tan(a+b)tan(a-b)]
=[1/2-1/3]/[1+1/2*1/3]
=[3-2]/(6+1)
=1/7.