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Very urgent.... In a triangle if tan A tanB tanC are in hp then fimd min value of cotB/2

Ani , 8 Years ago
Grade 11
anser 1 Answers
Vikas TU

Last Activity: 8 Years ago

if tanA tanB and tanC are in HP,
then,
2/tanB = 1/tanA + 1/tanC
1/2CotB = 1/(4tanA) + 1/(4tanC)
Z (let) = 1/(4tanA) + 1/(4tanC) …................(1)
 
Now, In a triangle,
A + B + C =180
A + C = 180 – B
tan(A+C) = tan(180 – B)
(tanA + tanC)/(1 – tanAtanC) = -tanB
 

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