Show that #(a^2sin(B-C))/(sinB+sinC)+(b^2sin(C-A))/(sinC+sinA)+(c^2sin(A-B))/(sinA+sinB)=0#?

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Answer 1 · 2018-01-16T16:29:53

1st part

#(a^2sin(B-C))/(sinB+sinC)#

#=(4R^2sinAsin(B-C))/(sinB+sinC)#

#=(4R^2sin(pi-(B+C))sin(B-C))/(sinB+sinC)#

#=(4R^2sin(B+C)sin(B-C))/(sinB+sinC)#

#=(4R^2(sin^2B-sin^2C))/(sinB+sinC)#

#=4R^2(sinB-sinC)#

Similarly

2nd part

#=(b^2sin(C-A))/(sinC+sinA)#

#=4R^2(sinC-sinA)#

3rd part

#=(c^2sin(A-B))/(sinA+sinB)#

#=4R^2(sinA-sinB)#

Adding three parts we have

The given expression #=0#