Question #991fa

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Answer 1 · 2017-11-22T01:08:51

#LHS=Sin(a+b)sin(a-b)+sin(b-c)sin(b+c)+sin(c-a)sin(c+a)#

#=1/2(2Sin(a+b)sin(a-b)+2sin(b-c)sin(b+c)+2sin(c-a)sin(c+a))#

#=1/2(cos2b-cos2a+cos2c-cos2b+cos2a-cos2c)=0#

Allternative

#LHS=Sin(a+b)sin(a-b)+sin(b-c)sin(b+c)+sin(c-a)sin(c+a)#

#=Sin^2a-sin^2b+sin^2b-sin^2c+sin^2c-sin^2a#

#=0#