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In ΔABC, if cot A, cot B, cot C are in A.P. then show that a2, b2, c2 are also in A.P. - Mathematics and Statistics

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प्रश्न

In ΔABC, if cot A, cot B, cot C are in A.P. then show that a2, b2, c2 are also in A.P.

योग
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उत्तर

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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अध्याय 3: Trigonometric Functions - Exercise 3.2 [पृष्ठ ८८]

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