已知sin[A+B]=1/2,sin[A-B]=1/3.求tan[A+B]-tanA-tanB/tan^Btan[A+B].

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已知sin[A+B]=1/2,sin[A-B]=1/3.求tan[A+B]-tanA-tanB/tan^Btan[A+B].
已知sin[A+B]=1/2,sin[A-B]=1/3.求tan[A+B]-tanA-tanB/tan^Btan[A+B].

已知sin[A+B]=1/2,sin[A-B]=1/3.求tan[A+B]-tanA-tanB/tan^Btan[A+B].
sin(A+B)=sinAcosB+cosAsinB=1/2
sin(A-B)=sinAcosB-cosAsinB=1/3
sin(A+B)+sin(A-B)=2sinAcosB=5/6
sin(A+B)-sin(A-B)=2cosAsinB=1/6
因为 tan(A+B)=(tanA+tanB)/(1-tanAtanB)
所以 tanA+tanB=tan(A+B)+tan(A+B)tanAtanB
所以[tan(A+B)-tanA-tanB]/[tan平方B·tan(A+B)
=tanA/tanB=sinAcosB/cosAsinB=5

sina cosb + cosa sinb = 1/2
sina cosb - cosa sinb = 1/3
相加：
2 sina cosb = 5/6
sina cosb = 5/12
相减：
2 cosa sinb = 1/6
cosa sinb = 1/12
则：
sina cosb /(cosa sinb)=5

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sina cosb + cosa sinb = 1/2
sina cosb - cosa sinb = 1/3
相加：
2 sina cosb = 5/6
sina cosb = 5/12
相减：
2 cosa sinb = 1/6
cosa sinb = 1/12
则：
sina cosb /(cosa sinb)=5
sina /cosa = 5 sinb / cosb
所以tan a = 5 tan b
tan(a+b)=[tan a + tan b]/[1- tan a tan b]
= (5tanb+tanb)/(1-5tanb tanb)
=6tanb /(1-5tanbtanb)
[tan(a+b)-tana-tanb]/[tanb tanb tan(a+b)]
= [6tanb/(1-5tanbtanb) - 5tanb - tanb]/[tanb tanb 6 tanb/(1-5tanbtanb)]
分子分母乘以(1-5tanbtanb)
= [6tanb - 6tanb(1-5tanbtanb)]/[6tanbtanbtanb]
= [30tanbtanbtanb]/[6tanbtanbtanb]
=5

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