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If A+B+C=π, then find the minimum value of cot^2A+cot^2B+cot^2C. 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
We know that square of real numbers is always non-negative,Hence (cotA-cotB)²+(cotB-cotC)²+(cotC-cotA)²≥0⇒2(cot²A+cot²B+cot²C)-2(cotA·cotB+cotB·cotC+cotC·cotA)≥0⇒(cot²A+cot²B+cot²C)≥(cotA·cotB+cotB·cotC+cotC·cotA)We also know the trigonometric identity in a triangle that if Α+Β+C=π, then cotA·cotB+cotB·cotC+cotC·cotA=1,∴,(cot²A+cot²B+cot²C)≥1.Therefore, the minimum value of the given expression is 1. The equality holds when (cot²A+cot²B+cot²C)=1⇒(cotA-cotB)²+(cotB-cotC)²+(cotC-cotA)²=0 (Carrying out calculations identical to above, but in reverse order)⇒cotA=cotB=cotC (∵ squares are always non-negative, so if a sum of squares of real numbers is zero, then each term should be zero individually)∵ cotA=cotB=cotC and Α+Β+C=π, ∴ A=B=C=π/3Hence, the equality holds for equilateral triangle.
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