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Question
In ΔABC, if cot A, cot B, cot C are in A.P. then show that a2, b2, c2 are also in A.P.
Solution
By the sine rule,
`sin A/a = sin B/b = sin C/c` = k
∴ sin A = ka, sin B = kb, sin C = kc ...(i)
Now, cot A, cot B, cot C are in A.P.
∴ cot C – cot B = cot B – cot A
∴ cot A + cot C = 2 cot B
∴ `cosA/sinA + cosC/sinC` = 2 cot B
∴ `(sinC cosA + sinA cosC)/(sinA. sinC)` = 2 cot B
∴ `(sin(A + C))/(sinA. sinC)` = 2 cot B
∴ `(sin(π - B))/(sinA. sinC)` = 2 cot B ...[∵ A + B + C = π]
∴ `sinB/(sinA. sinC) = (2cosB)/sinB`
∴ `sin^2 B/(sinA. sinC)` = 2 cos B
∴ `(k^2b^2)/((ka)(kc)) = 2((a^2 + c^2 - b^2)/(2ac))`
∴ `b^2/(ac) = (a^2 + c^2 - b^2)/(ac)`
∴ b2 = a2 + c2 – b2
∴ 2b2 = a2 + c2
Hence, a2 b2, c2 are in A.P.
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