1. Use the identity \(\cot 2\theta = \frac{1}{\tan 2\theta}\) and \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\) to express \(\cot 2\theta\) in terms of \(\tan \theta\):

\(\cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan \theta}\)

2. Use the identity \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\) to express \(\tan(\theta + 45^\circ)\):

\(\tan(\theta + 45^\circ) = \frac{\tan \theta + 1}{1 - \tan \theta}\)

3. Substitute these into the original equation:

\(\tan \theta \cdot \frac{\tan \theta + 1}{1 - \tan \theta} = 2 \cdot \frac{1 - \tan^2 \theta}{2 \tan \theta}\)

4. Simplify and rearrange to form a quadratic equation:

\(2 \tan^2 \theta + \tan \theta - 1 = 0\)

5. Solve the quadratic equation using the quadratic formula \(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = 1, c = -1\):

\(\tan \theta = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\)

6. This gives \(\tan \theta = \frac{1}{2}\) or \(\tan \theta = -1\).

7. Since \(0^\circ < \theta < 90^\circ\), \(\tan \theta = \frac{1}{2}\) is valid.

8. Therefore, \(\theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.6^\circ\).