Prove that sin^2a+sin^2b-sin^2c=2sinasinbsinc

= > (i) B + C = 180 - A

= > (ii) A = 180 - (B + C) = B + C

Now,

Sin^2A + sin^2B - sin^2C

= > sin^2A + sin(B + C)sin(B - C)

= > sin^2A + sin(180 - A) sin(B - C)

= > sin^2A + sinA sin(B - C)

= > sinA(sinA + sin(B - C))

= > sinA(sin(B + C) + sin(B - C)) [ from (ii)]

We know that sin(A + B) + sin(A - B) = 2 sinAcosB

= > sinA(2sinB cosc)

= > 2sinAsinBcosC.