We will simplify and verify the given equation step by step.

Given equation:

$\tan \left(\frac{\pi}{4} - \theta\right) = \frac{\cos\theta \sin\theta}{\cos\theta \sin\theta}$

Since the right-hand side simplifies to 1, this equation reduces to: $\tan \left(\frac{\pi}{4} - \theta\right) = 1$

Now, using the tangent subtraction formula: $\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Let $A = \frac{\pi}{4}$ and $B = \theta$, so: $\tan \left(\frac{\pi}{4} - \theta\right) = \frac{\tan \frac{\pi}{4} - \tan \theta}{1 + \tan \frac{\pi}{4} \tan \theta}$

Since $\tan \frac{\pi}{4} = 1$, we substitute:

$\tan \left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta}$

Thus, we get: $\frac{1 - \tan \theta}{1 + \tan \theta} = 1$

For this equation to hold true: $1 - \tan \theta = 1 + \tan \theta$

which simplifies to: $- \tan \theta = \tan \theta$

This implies: $\tan \theta = 0$

which happens when: $\theta = k\pi, \quad k \in \mathbb{Z}$

Conclusion:

The equation holds only for $\theta = k\pi$, where $k$ is any integer.

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Would you like a more detailed breakdown? Let me know! Here are some related questions to explore:

1. What is the general formula for $\tan(A - B)$?
2. How does the tangent function behave at different quadrants?
3. Why is $\tan \frac{\pi}{4} = 1$?
4. What happens when $\tan \theta = 0$, and how do we find such angles?
5. How can we prove the identity using sine and cosine instead of tangent?

Tip: The identity $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ is useful in trigonometry, especially in transformations and proofs.