Question

已知tan(a-b)=1/3,tan(a+b)=1/2,求tan2a,tan2b的值

Answer 1

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