State the Sine Rule for a triagle.

Prove, in the usual notation for a triangle ABC.

`(a^2-b^2)/c^2 = (sin(A-B))/(sin(A+B))`

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In a triangle

`sin(A)/a=sin(B)/b=sin(C)/c`

This is known as Sine rule.

LHS=`(a^2-b^2)/c^`

`=(sin^2(A)-sin^2(B))/(sin^2(C))`

`=((sin(A)-sin(B))(sin(A)+sin(B)))/(sin^2(C))`

`=((2sin((A+B)/2)cos((A-B)/2))(2cos((A+B)/2)sin((A-B)/2)))/(sin^2(C))`

`=(sin(A+B)sin(A-B))/(sin^2(C))`

`A+B+C=180^0`

`A+B=180^o-C`

`sin(A+B)=sin(180^o-C)=sin(C)`

`therefore`

`LHS=(sin(C)sin(A-B))/(sin^2(C))`

`=sin(A-B)/sin(C)`

`` =`sin(A-B)/sin(A+B)`

`=RHS`

Hence proved.

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