Question

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Prove:
sin (A-B) sin (A+B) = sin^2 A - sin^2 B

Answer 1

sin(A+B) sin(A -B)
= (sinA cosB + cosA sinB) (sinA cosB - cosA sinB)
= sin^2A cos^2B - cos^2A sin^2B
= sin^2A (1-sin^2B) - cos^2A sin^2B
= sin^2A - sin^2B (sin^2A + cos^2A)
= sin^2A - sin^2B.

Answer 2

sin (A-B) sin (A+B) = sin^2 A - sin^2 B
LHS=sin (A-B) sin (A+B)
=[sinAcosB-cosAsinB][sinAcosB+cosAsinB]
\[=\sin^2A \cos^2B-\cos^2A \sin^2B\]
=sin ² A ( 1 - sin ² B ) - sin ² B ( 1 - sin ² A )

=sin ² A - sin ² B - sin ² A sin ² B + sin ² A sin ² B

=sin ² A - sin ² B = RHS
Hence proved

Answer 3

Thank you!!