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Grade: 11

                        

If A+B+C=π, then find the minimum value of cot^2A+cot^2B+cot^2C.

2 years ago

Answers : (1)

Arun
25211 Points
							
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
2 years ago
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