Question

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

Answer 1

{sin(A+B)+sin(A-B)}×{sin(A+B)-sin(A-B)}
=2sinAcosB×2cosAsinB
=2sinAcos×2sinBcosB
=sin2A sin2B
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Answer 2

Answer:

It can be proved by using identities.

Step-by-step explanation:

We need to prove that

$sin^2 (A+B) - sin^2 (A-B) = sin 2A sin 2B$

Taking LHS

$sin^2 (A+B) - sin^2 (A-B)$

$={sin(A+B)+sin(A-B)}*{sin(A+B)-sin(A-B)}$

$=2sinAcosB*2cosAsinB$

$=2sinAcosA*2sinBcosB$

$=sin2Asin2B$

=RHS

Hence, LHS=RHS.

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