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Question
In Δ ABC, if a, b, c are in A.P., then show that cot `"A"/2, cot "B"/2, cot "C"/2` are also in A.P.
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Solution
a, b, c are in A.P.
∴ 2b = a + c ....(1)
Now,
`cot "A"/2 + cot "C"/2`
`= (cos "A"/2)/(sin "A"/2) + (cos "C"/2)/(sin "C"/2)`
`= (cos "A"/2 . sin "C"/2 + sin "A"/2. cos "C"/2)/(sin "A"/2. sin "C"/2)`
`= (sin ("A"/2 + "C"/2))/(sin "A"/2. sin "C"/2)`
`= (sin (pi/2 - "B"/2))/(sqrt((("s - b")("s - c"))/"bc"). sqrt((("s - a")("s - b"))/"ab"))` .....[∵ A + B + C = π]
`= (cos "B"/2)/((("s - b")/"b"). sqrt((("s - c")("s - a"))/"ca")`
`= ("b cos" "B"/2)/(("s - b"). sin "B"/2)`
`= "b"/("s - b"). cot "B"/2`
`= "b"/((("a + b + c")/2 - "b")). cot "B"/2` ....[∵ 2s = a + b + c]
`= ("2b"/("a + c - b")).cot "B"/2`
`= "2b"/(("2b - b")). cot "B"/2` ....[By (1)]
`= "2b"/"b".cot "B"/2`
∴ `cot "A"/2 + cot "C"/2 = 2 cot "B"/2`
Hence, `cot "A"/2, cot "B"/2, cot "C"/2` are in A.P.
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