ABCD

Since $ABCD$ is inscribed in a circle, thus $\angle BAD = 180^\circ - \angle BCD$ . So $\cos \angle BAD = \cos (180^\circ - \angle BCD) = -\cos \angle BCD$ .

Applying the cosine rule on triangle $ABD$ , we have

$\begin{aligned} BD^2 &= AB^2 + AD^2 - 2(AB)(AD)\cos \angle BAD \\ &= 1 + 36 - 2(1)(6)\cos \angle BAD \\ &= 37 - 12\cos \angle BAD \\ \end{aligned}$

Applying the cosine rule on triangle $BCD$ , we have

$\begin{aligned} BD^2 &= BC^2 + DC^2 - 2(BC)(CD) \cos \angle BCD \\ &= 9 + 16 + 2(3)(4) \cos \angle BAD \\ &= 25 + 24\cos \angle BAD\\ \end{aligned}$

Equating the two equations, we have $37 - 12\cos \angle BAD = 25 + 24 \cos \angle BAD$ , thus $\cos \angle BAD = \frac{37-25}{12+24} = \frac{1}{3}$ . Hence $\sec^2 \angle BAD = \frac{1}{\cos^2 \angle BAD} = \frac{1}{\frac{1}{9}} = 9$ .