1. The problem states a triangle $PQR$ with angles $\alpha$, $\beta$, and $\gamma$ such that $\alpha + \beta + \gamma = \pi$. 2. We need to verify or simplify the expression: $\sin^2 \alpha - \sin^2 \beta + \sin^2 \gamma = 2 \sin \alpha \cos \beta \sin \gamma$. 3. Use the Pythagorean sine identity difference: $\sin^2 \alpha - \sin^2 \beta = (\sin \alpha - \sin \beta)(\sin \alpha + \sin \beta)$. 4. Also recall the sine addition formulas: - $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ - $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$ 5. Since $\alpha + \beta + \gamma = \pi$, then $\gamma = \pi - (\alpha + \beta)$. 6. We know $\sin \gamma = \sin(\pi - (\alpha + \beta)) = \sin(\alpha + \beta)$. 7. Substitute $\sin \gamma$ in the right-hand side (RHS): $$2 \sin \alpha \cos \beta \sin \gamma = 2 \sin \alpha \cos \beta \sin(\alpha + \beta)$$ 8. Rewrite RHS using the sum formula: $$= 2 \sin \alpha \cos \beta (\sin \alpha \cos \beta + \cos \alpha \sin \beta) = 2 \sin \alpha \cos \beta \sin \alpha \cos \beta + 2 \sin \alpha \cos \beta \cos \alpha \sin \beta$$ 9. Simplify terms: $$= 2 \sin^2 \alpha \cos^2 \beta + 2 \sin \alpha \cos \alpha \cos \beta \sin \beta$$ 10. Now recollect the left-hand side (LHS): $\sin^2 \alpha - \sin^2 \beta + \sin^2 \gamma$. 11. Substitute $\sin^2 \gamma = \sin^2(\alpha + \beta)$. 12. Expand $\sin^2(\alpha + \beta) = (\sin \alpha \cos \beta + \cos \alpha \sin \beta)^2 = \sin^2 \alpha \cos^2 \beta + 2 \sin \alpha \cos \alpha \sin \beta \cos \beta + \cos^2 \alpha \sin^2 \beta$. 13. Thus LHS becomes: $$\sin^2 \alpha - \sin^2 \beta + \sin^2 \alpha \cos^2 \beta + 2 \sin \alpha \cos \alpha \sin \beta \cos \beta + \cos^2 \alpha \sin^2 \beta$$ 14. Group terms: $$= \sin^2 \alpha (1 + \cos^2 \beta) + \sin^2 \beta (\cos^2 \alpha - 1) + 2 \sin \alpha \cos \alpha \sin \beta \cos \beta$$ 15. Use $\cos^2 x = 1 - \sin^2 x$ to simplify: $$1 + \cos^2 \beta = 1 + 1 - \sin^2 \beta = 2 - \sin^2 \beta$$ $$\cos^2 \alpha - 1 = (1 - \sin^2 \alpha) - 1 = -\sin^2 \alpha$$ 16. Substitute in the expression: $$= \sin^2 \alpha (2 - \sin^2 \beta) - \sin^2 \beta \sin^2 \alpha + 2 \sin \alpha \cos \alpha \sin \beta \cos \beta$$ 17. Simplify: $$= 2 \sin^2 \alpha - \sin^2 \alpha \sin^2 \beta - \sin^2 \beta \sin^2 \alpha + 2 \sin \alpha \cos \alpha \sin \beta \cos \beta$$ $$= 2 \sin^2 \alpha - 2 \sin^2 \alpha \sin^2 \beta + 2 \sin \alpha \cos \alpha \sin \beta \cos \beta$$ 18. Factor out 2: $$= 2 \left( \sin^2 \alpha - \sin^2 \alpha \sin^2 \beta + \sin \alpha \cos \alpha \sin \beta \cos \beta \right)$$ 19. This equals the expanded RHS from step 9. 20. Therefore, the expression holds true by trigonometric identities and the angle sum property. **Final answer: The given identity $\sin^2 \alpha - \sin^2 \beta + \sin^2 \gamma = 2 \sin \alpha \cos \beta \sin \gamma$ is verified using the triangle angle sum and trigonometric identities.**