in triangle ABC, if angle C= pi/2, then prove that sin (A+B)= a2-b2/a2+b2 - Maths -

Question

in triangle ABC, if angle C= pi/2, then prove that sin (A+B)=
a2-b2/a2+b2

Anuradha Sharma · Accepted Answer

Hi,

taking RHS, a2-b2a2+b2= k2sin2A-k2sin2Bk2sin2A+k2sin2B  using sine rule asinA=bsinB=csinC=k= sin2A-sin2Bsin2A+sin2Bsince C  = 90 so A+B= 90 sin2A+sin2B= sin2A+sin290-A= sin2A+cos2A= 1 sin2A-sin2Bsin2A+sin2B=sin2A-sin2B=sin2A-cos2A= -cos2ABut nowhere we are getting RHS as sinA+Bso please recheck the question
Regards