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Question
In ΔABC, prove that `tan((A - B)/2) = (a - b)/(a + b)*cot C/2`.
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Solution
By sine rule, `a/(sin A) = b/(sin B) = c/(sin C) = k`
∴ a = k sin A, b = k sin B, c = k sin C
RHS = `((a - b)/(a + b)) cot (C/2)`
= `((k sin A - k sin B)/(k sin A + k sin B)) cot(C/2)`
= `((sin A - sin B)/(sin A + sin B)) cot (C/2)`
= `(2 cos ((A + B)/2)*sin((A - B)/2))/(2 sin ((A + B)/2)*cos((A - B)/2)) xx (cos(C/2))/(sin(C/2))`
= `(cos(pi/2 - C/2)*sin((A - B)/2))/(sin(pi/2 - C/2)*cos((A - B)/2)) xx (cos (C/2))/(sin(C/2))` ...[∵A + B + C = π]
= `(sin(C/2))/(cos(C/2)) xx tan ((A - B)/2) xx (cos (C/2))/(sin(C/2))`
= `tan ((A - B)/2)` = LHS
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