1. Problem: Prove that $\frac{\sec \theta + \csc \theta \tan \theta}{\sec^2 \theta} = \cot \theta \csc \theta$. 2. Rewrite the left-hand side in terms of sine and cosine to simplify. $$\frac{\sec \theta + \csc \theta \tan \theta}{\sec^2 \theta}$$ 3. Use the identities $\sec \theta=\frac{1}{\cos \theta}$, $\csc \theta=\frac{1}{\sin \theta}$, and $\tan \theta=\frac{\sin \theta}{\cos \theta}$. $$\frac{\frac{1}{\cos \theta} + \frac{1}{\sin \theta}\cdot\frac{\sin \theta}{\cos \theta}}{\left(\frac{1}{\cos \theta}\right)^2}$$ 4. Simplify the product in the numerator and combine like terms. $$\frac{\frac{1}{\cos \theta} + \frac{1}{\cos \theta}}{\frac{1}{\cos^2 \theta}}$$ 5. Combine the numerator to get $\frac{2}{\cos \theta}$ and then divide by $\frac{1}{\cos^2 \theta}$. $$\frac{\frac{2}{\cos \theta}}{\frac{1}{\cos^2 \theta}}$$ 6. Dividing by $\frac{1}{\cos^2 \theta}$ is the same as multiplying by $\cos^2 \theta$, which gives $2\cos \theta$. 7. Compute the right-hand side in sine and cosine to compare. $$\cot \theta \csc \theta = \frac{\cos \theta}{\sin \theta}\cdot\frac{1}{\sin \theta} = \frac{\cos \theta}{\sin^2 \theta}$$ 8. Compare the two sides and determine if the equality holds for all $\theta$. Since the left-hand side simplifies to $2\cos \theta$ and the right-hand side simplifies to $\frac{\cos \theta}{\sin^2 \theta}$, these are not equal for general $\theta$. 9. Solve for when they are equal by setting $2\cos \theta=\frac{\cos \theta}{\sin^2 \theta}$ and assuming $\cos \theta\neq 0$. Dividing both sides by $\cos \theta$ gives $2=\frac{1}{\sin^2 \theta}$, so $\sin^2 \theta=\frac{1}{2}$. 10. Therefore the equality holds only when $\sin^2 \theta=\frac{1}{2}$, which occurs at $\theta=\frac{\pi}{4}+k\frac{\pi}{2}$ for integer $k$, subject to the domain restrictions $\sin \theta\neq 0$ and $\cos \theta\neq 0$ so the original expressions are defined. Final answer: The identity is false in general; it holds exactly for $\theta=\frac{\pi}{4}+k\frac{\pi}{2}$ with $k\in\mathbb{Z}$.