Trig Identity Cea6B5

1. **Problem Statement:** Show that $$\sec^4 \theta (1 - \sin^4 \theta) - 2 \tan^2 \theta = 1$$. 2. **Recall the identities:** - $$\sec \theta = \frac{1}{\cos \theta}$$ - $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ - $$1 - \sin^2 \theta = \cos^2 \theta$$ 3. **Rewrite the expression:** $$\sec^4 \theta (1 - \sin^4 \theta) - 2 \tan^2 \theta = ?$$ Note that $$1 - \sin^4 \theta = (1 - \sin^2 \theta)(1 + \sin^2 \theta) = \cos^2 \theta (1 + \sin^2 \theta)$$. 4. **Substitute:** $$\sec^4 \theta (1 - \sin^4 \theta) = \sec^4 \theta \cdot \cos^2 \theta (1 + \sin^2 \theta) = \sec^4 \theta \cos^2 \theta (1 + \sin^2 \theta)$$ Since $$\sec \theta = \frac{1}{\cos \theta}$$, then $$\sec^4 \theta = \frac{1}{\cos^4 \theta}$$. So, $$\sec^4 \theta \cos^2 \theta (1 + \sin^2 \theta) = \frac{1}{\cos^4 \theta} \cdot \cos^2 \theta (1 + \sin^2 \theta) = \frac{\cos^2 \theta}{\cos^4 \theta} (1 + \sin^2 \theta) = \frac{1}{\cos^2 \theta} (1 + \sin^2 \theta) = \sec^2 \theta (1 + \sin^2 \theta)$$ 5. **Rewrite the original expression:** $$\sec^2 \theta (1 + \sin^2 \theta) - 2 \tan^2 \theta$$ 6. **Express $$\tan^2 \theta$$ in terms of sine and cosine:** $$\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = \sin^2 \theta \sec^2 \theta$$ 7. **Substitute:** $$\sec^2 \theta (1 + \sin^2 \theta) - 2 \sin^2 \theta \sec^2 \theta = \sec^2 \theta (1 + \sin^2 \theta - 2 \sin^2 \theta) = \sec^2 \theta (1 - \sin^2 \theta)$$ 8. **Recall that $$1 - \sin^2 \theta = \cos^2 \theta$$:** $$\sec^2 \theta \cos^2 \theta = 1$$ 9. **Final answer:** $$\boxed{1}$$ Thus, the given expression simplifies to 1 as required.