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.