Step 1: Given condition
We are given $\sin \theta = \cos \theta$.
Step 2: Divide both sides by $\cos \theta$ (valid for $0^\circ \leq \theta \leq 90^\circ$, except $\theta=90^\circ$)
\[ \frac{\sin \theta}{\cos \theta} = \frac{\cos \theta}{\cos \theta} \] \[ \tan \theta = 1 \]
Step 3: Solve for $\theta$
\[ \tan \theta = 1 \,\Rightarrow\ \theta = 45^\circ \]
Step 4: Check interval
Since $0^\circ \leq \theta \leq 90^\circ$, the only solution is $\theta = 45^\circ$.
\[ \boxed{\theta = 45^\circ} \]
The value of $\dfrac{1+\cot^2 \theta}{1+\tan^2 \theta}$ will be: