Show that #(b^2-c^2)*cotA+(c^2-a^2)*cotB+(a^2-b^2)*cotC=0#?

1 Answer
Jan 16, 2018

By sine law we know

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

Now
1st 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)#

Similarly

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

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

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

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

Adding three parts we get

Whole expression

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