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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.

Grade:11

2 Answers

Arun
25750 Points
5 years ago
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
Gaurav Kasliwal
13 Points
3 years ago
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/3
Hence, the equality holds for equilateral triangle.

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