已知tan(A-B⼀2)=2,tan(B-A⼀2)=-3 求:(1) tan[(A+B)⼀2] (2) tan(A+B)

2024-12-30 04:41:12
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回答1:

⑴tan[(A+B)/2]=tan[(A-B/2)+(B-A/2)]
=(tan(A-B/2)+tan(B-A/2))/(1-tan(A-B/2)·tan(B-A/2))
=(2-3)/1-(2·-3)=-1/7
⑵tan(A+B)=tan[2·(A+B)/2]
=(2tan(A+B)/2)/(1-(tan(A+B)/2)^2)
=2·(-1/7)/1-(-1/7)^2
=-7/24