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Question

If A, B, C are actual positive angles such that A + B + C = π and cot A cot B cot C = k then


A

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B

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C

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D

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Solution

The correct option is A


A + B + C =π

A + B = π - C

tan (A + B = tan (π - C)

tanA+tanB1tanA.tanB = -tanC

tan A + tan B + tan C = tan A tan B tan C - - - - - - (1)

For three values tan A, tan B & tan C

AM GM

tanA+tanB+tanC3 (tanA.tanB.tanC)13

[tanA + tanB + tanC = tanA tanB tanC]

(tanA.tanB.tanC)233

(1cotA.cotB.cotC)233

{cotA.cotB.cotC = k}

(1k)233

1k(3)32

k 1(3)32

k 13.312

k 133


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