Squaring the given equation

Given $\tan \theta + \cot \theta = 13$. Squaring both sides using $(a+b)^2 = a^2 + b^2 + 2ab$:

$(\tan \theta + \cot \theta)^2 = 13^2$

Expanding the expression

Expanding the left side and noting that $\tan \theta \cdot \cot \theta = 1$:

$\tan^2 \theta + \cot^2 \theta + 2(\tan \theta \cdot \cot \theta) = 169$

Simplifying

Since $\tan \theta \cdot \cot \theta = 1$, we get:

$\tan^2 \theta + \cot^2 \theta + 2 = 169 \implies \tan^2 \theta + \cot^2 \theta = 167$

Final Calculation

The question asks for $8(\tan^2 \theta + \cot^2 \theta)$:

$8 \times 167 = 1336$