1. **State the problem:** Verify if the identity $\sin(a-b)\sin(a-b) = \sin^2 a - \sin^2 b$ is true. 2. **Recall the formula:** The left side is $\sin^2(a-b)$. 3. **Use the sine difference formula:** $$\sin(a-b) = \sin a \cos b - \cos a \sin b$$ 4. **Square the expression:** $$\sin^2(a-b) = (\sin a \cos b - \cos a \sin b)^2 = \sin^2 a \cos^2 b - 2 \sin a \cos b \cos a \sin b + \cos^2 a \sin^2 b$$ 5. **Simplify the middle term:** $$-2 \sin a \cos b \cos a \sin b = -2 \sin a \cos a \sin b \cos b$$ 6. **Rewrite the right side:** $$\sin^2 a - \sin^2 b$$ 7. **Compare both sides:** The left side contains mixed terms and products of sines and cosines, while the right side is a difference of squares of sines. 8. **Conclusion:** The identity $\sin^2(a-b) = \sin^2 a - \sin^2 b$ is **not true** in general. **Final answer:** The given equation is false; $\sin^2(a-b) \neq \sin^2 a - \sin^2 b$ in general.