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Grade: 11
        In properties of triangles:-In ∆ABC 1+cos(A-B)cosC/1+cos(A-C)cosB = ?
10 months ago

Answers : (1)

Arun
14688 Points
							
Dear Rohan
 
2cos(A-B)cosC = cos(A-B+C) + cos(A-B-C) = cos(180-2B)+cos(2A-180) = - [cos2B+cos2A] 
In the second br B and C are interchanged so it is - [cos2C + cos2A] 
LHS = [1+cos(A-B) cosC] / [1+ cos(A-C) cosB] 
=[2+2cos(A-B) cosC] / [2+ 2cos(A-C) cosB] 
=[1 - cos2A +1 - cos2B]/[1 - cos2C +1 - cos2A] 
= [2sin^2A + 2sin^2B]/92sin^2C+2sin^2A]
(a^2 +b^2)/(a^2 + c^2) by using sine rule.
 
 
Regards
Arun (askIITians forum expert)
10 months ago
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