Given: $\tan\theta + \dfrac{1}{\tan\theta} = 2$
Squaring both sides:
$$\left(\tan\theta + \frac{1}{\tan\theta}\right)^2 = 4$$
$$\tan^2\theta + 2\cdot\tan\theta\cdot\frac{1}{\tan\theta} + \frac{1}{\tan^2\theta} = 4$$
$$\tan^2\theta + \frac{1}{\tan^2\theta} + 2 = 4$$
$$\therefore \tan^2\theta + \frac{1}{\tan^2\theta} = 2$$
The key step is squaring the given expression and using the identity $(a+b)^2 = a^2 + 2ab + b^2$. The middle term $2 \cdot \tan\theta \cdot \frac{1}{\tan\theta} = 2$ simplifies neatly. Examiners expect the squaring step to be shown clearly — don't skip it. This is a standard algebraic manipulation question disguised as trigonometry.