1. **State the problem:** We want to show that $$3 \tan^2 \theta + 5 \sin^2 \theta \equiv \frac{8 \sin^2 \theta - 5 \sin^4 \theta}{1 - \sin^2 \theta}$$ for any angle $$\theta$$. 2. **Recall the definitions and formulas:** - $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ - $$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}$$ - Also, $$\cos^2 \theta = 1 - \sin^2 \theta$$ (Pythagorean identity). 3. **Rewrite the left side using these formulas:** $$3 \tan^2 \theta + 5 \sin^2 \theta = 3 \frac{\sin^2 \theta}{\cos^2 \theta} + 5 \sin^2 \theta$$ 4. **Substitute $$\cos^2 \theta$$ with $$1 - \sin^2 \theta$$:** $$= 3 \frac{\sin^2 \theta}{1 - \sin^2 \theta} + 5 \sin^2 \theta$$ 5. **Get a common denominator to combine the terms:** Rewrite $$5 \sin^2 \theta$$ as $$\frac{5 \sin^2 \theta (1 - \sin^2 \theta)}{1 - \sin^2 \theta}$$: $$= \frac{3 \sin^2 \theta}{1 - \sin^2 \theta} + \frac{5 \sin^2 \theta (1 - \sin^2 \theta)}{1 - \sin^2 \theta}$$ 6. **Add the fractions:** $$= \frac{3 \sin^2 \theta + 5 \sin^2 \theta (1 - \sin^2 \theta)}{1 - \sin^2 \theta}$$ 7. **Expand the numerator:** $$= \frac{3 \sin^2 \theta + 5 \sin^2 \theta - 5 \sin^4 \theta}{1 - \sin^2 \theta}$$ 8. **Combine like terms in the numerator:** $$= \frac{8 \sin^2 \theta - 5 \sin^4 \theta}{1 - \sin^2 \theta}$$ 9. **Conclusion:** The left side simplifies exactly to the right side, so the identity is proven. This means $$3 \tan^2 \theta + 5 \sin^2 \theta$$ is equal to $$\frac{8 \sin^2 \theta - 5 \sin^4 \theta}{1 - \sin^2 \theta}$$ for all $$\theta$$ where the expressions are defined.