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Question
In ΔABC, prove that `tan((A - B)/2) = (a - b)/(a + b)*cot C/2`
Solution
In ΔABC by sine rule, we have
`a/sinA = b/sinB = c/sinC` = k
∴ a = k sin A, b = k sin B and c = k sin C
Now, consider
`(a - b)/(a + b) = (ksinA - ksinB)/(ksinA + ksinB)`
= `(sinA - sinB)/(sinA + sinB)`
= `(2cos((A + B)/2)sin((A - B)/2))/(2sin((A + B)/2)cos((A - B)/2))`
= `cot((A + B)/2)*tan((A - B)/2)`
= `cot(pi/2 - C/2).tan((A - B)/2)` .....[∵ A + B + C = π]
= `tan(C/2)tan((A - B)/2)`
∴ `(a - b)/(a + b) = tan C/2* tan((A - B)/2)`
∴ `tan((A - B)/2) = (a - b)/(a + b)cot(C/2)`
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