Step 1: Express \(\cot \theta\) and \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).

\(\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}\)

Step 2: Substitute into the equation:

\(\frac{\cos \theta}{\sin \theta} - 2 \left( \frac{\sin \theta}{\cos \theta} \right) = \sin 2\theta\)

Step 3: Use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\):

\(\frac{\cos^2 \theta - 2 \sin^2 \theta}{\sin \theta \cos \theta} = 2 \sin \theta \cos \theta\)

Step 4: Multiply through by \(\sin \theta \cos \theta\) to clear the fraction:

\(\cos^2 \theta - 2 \sin^2 \theta = 2 \sin^2 \theta \cos^2 \theta\)

Step 5: Use \(\sin^2 \theta = 1 - \cos^2 \theta\):

\(\cos^2 \theta - 2(1 - \cos^2 \theta) = 2 \cos^2 \theta (1 - \cos^2 \theta)\)

Step 6: Simplify:

\(\cos^2 \theta - 2 + 2 \cos^2 \theta = 2 \cos^2 \theta - 2 \cos^4 \theta\)

Step 7: Rearrange to form a quadratic in \(\cos^2 \theta\):

\(2 \cos^4 \theta + \cos^2 \theta - 2 = 0\)

Step 8: Solve for \(\cos^2 \theta\):

Let \(x = \cos^2 \theta\), then:

\(2x^2 + x - 2 = 0\)

Step 9: Solve the quadratic equation using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

For \(a = 2, b = 1, c = -2\):

\(x = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4}\)

Step 10: Find \(\theta\) for \(90^\circ < \theta < 180^\circ\):

Only the positive root is valid for \(\cos^2 \theta\), so:

\(\cos^2 \theta = \frac{-1 + \sqrt{17}}{4}\)

\(\theta = 152.1^\circ\)