If # alpha +beta -gamma =pi# Prove that #(sin^2alpha + sin^2beta - sin^2gamma)=#2sin alpha ×2sin beta ×2 cos gamma#?

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Answer 1 · 2018-07-09T03:09:41

#LHS=sin^2alpha+sin^2beta-sin^2gamma#

#=1/2(1-cos2alpha)+1/2(1-cos2beta)-1/2(1-cos2gamma)#

#=1/2(1+cos2gamma-(cos2alpha+cos2beta))#

#=1/2(2cos^2gamma-(2cos(alpha+beta)cos(alpha-beta))#

#=cos^2gamma-cos(pi+gamma)cos(alpha-beta)#
#=cosgamma*cosgamma+cosgammacos(alpha-beta)#

#=cosgamma*cosgamma+cosgammacos(alpha-beta)#

#=cosgamma(cos(alpha-beta)+cos(alpha+beta-pi))#

#=cosgamma(cos(alpha-beta)-cos(alpha+beta))#

#=cosgamma*2sinalphasinbeta#