Solution:

The equation is equivalent to $1+\sin \theta+\cos \theta=5 \sin \theta \cos \theta$. Let $A := \sin \theta + \cos \theta$ and $B := \sin \theta \cos \theta$. Then $1+A=5B$ and $A^{2}=1+2B$. As $1+A \neq 0$,
$$1+A=\frac{5\left(A^{2}-1\right)}{2} \Longrightarrow 1=\frac{5(A-1)}{2}$$
which gives $A=\frac{7}{5}$. Consequently, we get $B=\frac{12}{25}$. Hence,
$$(1+\tan \theta)(1+\cot \theta)=\frac{A^{2}}{B}=\frac{49}{25} \cdot \frac{25}{12}=\frac{49}{12}$$