1. The problem is to simplify or work with the expression involving $\tan^2 A$, not $\tan 2A$. 2. Recall the identity for $\tan^2 A$: it is simply the square of $\tan A$, i.e., $\tan^2 A = (\tan A)^2$. 3. If you need to express $\tan^2 A$ in terms of sine and cosine, use the definition $\tan A = \frac{\sin A}{\cos A}$, so $$\tan^2 A = \left(\frac{\sin A}{\cos A}\right)^2 = \frac{\sin^2 A}{\cos^2 A}.$$ 4. This is different from $\tan 2A$, which is the tangent of double angle $2A$ and has the formula $$\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}.$$ 5. Make sure to distinguish between $\tan^2 A$ (square of tangent) and $\tan 2A$ (tangent of double angle) when solving problems. Final answer: $\tan^2 A = \frac{\sin^2 A}{\cos^2 A}$.