Mostrar que (b ^ 2-c ^ 2) * cotA + (c ^ 2-a ^ 2) * cotB + (a ^ 2-b ^ 2) * cotC = 0?

Per llei sine sabem

# a / sinA = b / sinB = c / sinC = 2R #

Ara

1a part

# (b ^ 2-c ^ 2) cotA #

# = (4R ^ 2sin ^ 2B-4R ^ 2sin ^ 2C) cotA #

# = 4R ^ 2 (1/2 (1-cos2B) -1/2 (1-cos2C) cotA #

# = 4R ^ 2xx1 / 2 (cos2C-cos2B) cotA #

# = 2R ^ 2xx2sin (B + C) sin (B-C) cosA / sinA #

# = 4R ^ 2sin (pi-A) sin (B-C) cosA / sinA #

# = 4R ^ 2sinAsin (B-C) cosA / sinA #

# = 4R ^ 2sin (B-C) cosA #

# = 4R ^ 2 (sinBcosCcosA-cosBsinCcosA) #

De la mateixa manera

2a part # = (c ^ 2-a ^ 2) cotB #

# = 4R ^ 2 (sinCcosAcosB-cosCsinAcosB) #

3a part # = (a ^ 2-b ^ 2) cotC #

# = 4R ^ 2 (sinAcosBcosC-cosAsinBcosC) #

Afegint tres parts

Expressió sencera

# (b ^ 2-c ^ 2) cotA + (c ^ 2-a ^ 2) cotB + (a ^ 2-b ^ 2) cotC = 0 #