Question

已知tan（A+B）=3,tan（A-B）=5

Answer 1

tan2A=tan(A+B+A-B)={tan(A+B)+tan(A-B)}/{1-tan(A+B)*tan(A-B)}=-4/7;

tan2B=tan(A+B-A+B)={tan(A+B)-tan(A-B)}/{1+tan(A+B)*tan(A-B)}=-1/8.