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If A+B+C=π, then find the minimum value of cot^2A+cot^2B+cot^2C.
i) a + b + c = π; so (a + b) = π - c ==> cot(a + b) = cot(π - c) = -cot(c) ==> {cot(a)*cot(b) - 1}/{cot(a) + cot(b)} = -cot(c) Simplifying the above, cot(a)*cot(b) + cot(b)*cot(c) + cot(c)*cot(a) = 1 ---------- (1) ii) {cot(a) + cot(b) + cot(c)}² = cot²a + cot²b + cot²c + 2{cot(a)*cot(b) + cot(b)*cot(c) + cot(c)*cot(a)} Substituting the value from (1) above, {cot(a) + cot(b) + cot(c)}² = cot²a + cot²b + cot²c + 2 ==> cot²a + cot²b + cot²c = {cot(a) + cot(b) + cot(c)}² - 2 iii) Further, when a + b + c = π, minimum value of cot(a) + cot(b) + cot(c) = √3 [That is each of the angle is π/3] Hence, Minimum value of cot²a + cot²b + cot²c = 1
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