Question

sin^2A + sin^2B – sin^2C = 2 sinAsinBsin C 

Answer 1

sin(2A)+sin(2B)+sin(2C)

=2sin(C)cos(A−B)+2sin(C)cos

(C)

=2sin(C)(cos(A−B)+cos(C))

=2sin

(C)(cos(A−B)+cos(π−(A+B)))

=2sin(C)(cos(A−B)−cos(A+B))

=2sin(C)×2sin(A)sin(B)

=4sin(A)sin(B)sin(C)

Answer 2

In the question U forgot to write, A + B + C = 180 and It should be 2sinAsinBcosC.

= > (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.

Hope it helps!