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If x, y and z are proportional to the cosines of angles A, B and C of the triangle ABC respectively then prove that (y+z)tan((B-C)/2)cotA/2 + (z+x)tan((C-A)/2)cotB/2 + (x+y)tan((A-B)/2)cotC/2 = 0

If x, y and z are proportional to the cosines of angles A, B and C of the triangle ABC respectively then prove that 
(y+z)tan((B-C)/2)cotA/2 + (z+x)tan((C-A)/2)cotB/2 + (x+y)tan((A-B)/2)cotC/2 = 0
 

Grade:10

1 Answers

srikar
40 Points
6 years ago
take triangle ABC as equilateral triangle then 
(y+z)tan((B-C)/2)cotA/2 = 0
(z+x)tan((C-A)/2)cotB/2  =0
(x+y)tan((A-B)/2)cotC/2 = 0
So,
(y+z)tan((B-C)/2)cotA/2 + (z+x)tan((C-A)/2)cotB/2 + (x+y)tan((A-B)/2)cotC/2 = 0

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