In any ∆ ABC, the value of 2ac sin((A – B + C)/2) is (a) a² + b² – c²(b) c² + a² – b²(c) b² – c² – a

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[IIT 2000] In a triangle ABC prove that 2ac*sin [(A-B+C)/2] = square(a) + square(c) - square(b). [IIT 2000] In a triangle ABC prove that 2ac*sin [(A-B+C)/2] = square(a) + square(c) - square(b).
In a `Delta ABC, 2ac Sin ((A-B+C)/2)` is equal to In a `Delta ABC, 2ac Sin ((A-B+C)/2)` is equal to
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In `Delta ABC, abcsin(A/2)sin(B/2)sin(C/2)=` In `Delta ABC, abcsin(A/2)sin(B/2)sin(C/2)=`