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The expression you're dealing with is:

\[
\frac{\tan 2\theta}{\cos 2\theta (1 - \sin 2\theta)}
\]

We need to express this in terms of \(\tan \theta\). 

### Step-by-Step Solution:

1. **Recall the double angle identities** for tangent, sine, and cosine:
   - \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\)
   - \(\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\)
   - \(\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}\)

2. **Substitute** these identities into the original expression.

   The numerator becomes:
   \[
   \frac{\tan 2\theta}{\cos 2\theta (1 - \sin 2\theta)} = \frac{\frac{2 \tan \theta}{1 - \tan^2 \theta}}{\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \left( 1 - \frac{2 \tan \theta}{1 + \tan^2 \theta} \right)}
   \]

3. Simplify the denominator:
   \[
   1 - \sin 2\theta = 1 - \frac{2 \tan \theta}{1 + \tan^2 \theta}
   \]

   Combine the terms in the denominator:
   \[
   1 - \frac{2 \tan \theta}{1 + \tan^2 \theta} = \frac{(1 + \tan^2 \theta) - 2 \tan \theta}{1 + \tan^2 \theta}
   \]
   So the denominator becomes:
   \[
   \cos 2\theta \left(1 - \sin 2\theta\right) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \cdot \frac{(1 + \tan^2 \theta) - 2 \tan \theta}{1 + \tan^2 \theta}
   \]

4. **Simplify the full expression**:
   The expression becomes:
   \[
   \frac{\frac{2 \tan \theta}{1 - \tan^2 \theta}}{\frac{1 - \tan^2 \theta}{(1 + \tan^2 \theta)^2} \left( (1 + \tan^2 \theta) - 2 \tan \theta \right)}
   \]

   Multiply by the reciprocal of the denominator:
   \[
   = \frac{2 \tan \theta (1 + \tan^2 \theta)^2}{(1 - \tan^2 \theta)^2 \left( (1 + \tan^2 \theta) - 2 \tan \theta \right)}
   \]

Thus, the expression in terms of \(\tan \theta\) is:

\[
\frac{2 \tan \theta (1 + \tan^2 \theta)^2}{(1 - \tan^2 \theta)^2 \left( (1 + \tan^2 \theta) - 2 \tan \theta \right)}
\]

Let me know if you'd like further steps or if anything needs clarification!

Here are 5 related questions for further practice:
1. What is the double angle formula for \(\sin 2\theta\) in terms of \(\tan \theta\)?
2. How do you derive the triple angle identity for \(\tan 3\theta\)?
3. How would you simplify \(\cos^2 \theta - \sin^2 \theta\) using the tangent identity?
4. Can you express \(\tan(45^\circ + \theta)\) in terms of \(\tan \theta\)?
5. What is the general approach to converting higher-angle trigonometric expressions into basic angles?

### Tip:
Whenever simplifying complex trigonometric expressions, it's helpful to remember key identities for sine, cosine, and tangent, especially the double and half-angle formulas.

Write in terms of tanθ of the following double-angled trigonometric expression: tan(2θ) / [cos(2θ)(1 - sin(2θ))].

Solve a double angle identity problem by expressing tan(2θ) / cos(2θ)(1 - sin(2θ)) in terms of tanθ. This step-by-step solution involves key trigonometric formulas for sine, cosine, and tangent, making it suitable for students in Grades 11-12.

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