Question

在锐角△ABC中,sin(A+B)=3/5,sin(A-B)=5/13则tan2B=

Answer 1

锐角△ABC中
∴A+B>90°
A-B<90°
sin(A+B)=3/5,sin(A-B)=5/13
∴cos(A+B)=-4/5
cos(A-B)=12/13
∴tan(A+B)=-3/4
tan(A-B)=5/12
tan(2B)
=tan[(A+B)-(A-B)]
=[tan(A+B)-tan(A-B)]/[1+tan(A+B)tan(A-B)]
=(-3/4-5/12)/(1-3/4*5/12)
=-56/33

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