tan(A+B)=3,tan(A-B)=5,求tan2A,tan2B.(A,B是角)

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tan2A=tan[(A+B)+(A-B)]
=[tan(A+B)+tan(A-B)]/[1-tan(A+B)tan(A-B)]
=(3+5)/(1-3×5）
=-4/7tan2B=tan[(A+B)-(A-B)]
=[tan(A+B)-tan(A-B)]/[1+tan(A+B)tan(A-B)]
=(3-5)/(1+3×5）
=-1/8======以下答案可供参考======
供参考答案1:
tan2A=tan[（A+B）+（A-B）]=（3+5）/（1-15）=-4/7
tan2B=tan[（A+B）-（A-B）]=（3-5）/（1+15）谢谢
=-1/8供参考答案2:
tan2A=tan[(A+B)+(A-B)]
=(3+5)/(1-3*5)
=-4/7.
tan2B=tan[(A+B)-(A-B)]
=(3-5)/(1+3*5)
=-1/8.
供参考答案3:
tan2A=tan[(A+B)+(A-B)]
=[tan(A+B)+tan(A-B)]/[1-tan(A+B)tan(A-B)]=-4/7tan2B=tan[(A+B)-(A-B)]
=[tan(A+B)-tan(A-B)]/[1+tan(A+B)tan(A-B)]=-1/8