1. **State the problem:** Prove the identity $$\sec \Theta + \csc \Theta \tan \Theta = \cot \Theta \csc \Theta.$$\n\n2. **Write each trigonometric function in terms of sine and cosine:**\n$$\sec \Theta = \frac{1}{\cos \Theta}, \quad \csc \Theta = \frac{1}{\sin \Theta}, \quad \tan \Theta = \frac{\sin \Theta}{\cos \Theta}, \quad \cot \Theta = \frac{\cos \Theta}{\sin \Theta}.$$\n\n3. **Substitute these into the left-hand side (LHS):**\n$$\text{LHS} = \frac{1}{\cos \Theta} + \left( \frac{1}{\sin \Theta} \right) \left( \frac{\sin \Theta}{\cos \Theta} \right).$$\n\n4. **Simplify the second term:**\n$$\frac{1}{\sin \Theta} \times \frac{\sin \Theta}{\cos \Theta} = \frac{1}{\cos \Theta}.$$\n\n5. **So the LHS becomes:**\n$$\frac{1}{\cos \Theta} + \frac{1}{\cos \Theta} = \frac{2}{\cos \Theta}.$$\n\n6. **Now simplify the right-hand side (RHS):**\n$$\text{RHS} = \cot \Theta \csc \Theta = \left( \frac{\cos \Theta}{\sin \Theta} \right) \left( \frac{1}{\sin \Theta} \right) = \frac{\cos \Theta}{\sin^2 \Theta}.$$\n\n7. **We see that LHS $$\frac{2}{\cos \Theta}$$ and RHS $$\frac{\cos \Theta}{\sin^2 \Theta}$$ are not obviously equal. Let's re-express the problem:**\n\nPerhaps the problem meant to prove a different identity. Let's re-express original expression carefully:$$\sec \Theta + \csc \Theta \tan \Theta = \cot \Theta \csc \Theta.$$\n\n8. **Try to manipulate the RHS to match terms on the LHS instead:**\nRewrite RHS:\n$$\cot \Theta \csc \Theta = \frac{\cos \Theta}{\sin \Theta} \cdot \frac{1}{\sin \Theta} = \frac{\cos \Theta}{\sin^2 \Theta}.$$\n\n9. **Try to express the LHS as a single fraction:**\n$$\sec \Theta + \csc \Theta \tan \Theta = \frac{1}{\cos \Theta} + \frac{1}{\sin \Theta} \cdot \frac{\sin \Theta}{\cos \Theta} = \frac{1}{\cos \Theta} + \frac{1}{\cos \Theta} = \frac{2}{\cos \Theta}.$$\n\n10. **This means LHS equals $$\frac{2}{\cos \Theta}$$ but RHS equals $$\frac{\cos \Theta}{\sin^2 \Theta}$$. They are only equal if $$\frac{2}{\cos \Theta} = \frac{\cos \Theta}{\sin^2 \Theta}$$, which is not generally true. Therefore, the identity is false as stated.\n\n11. **Conclusion:** The given equation is not a true identity unless restrictions or corrections are made. Please verify the original problem statement or provide additional context.