[proofplan] We first prove the continuant determinant identity for consecutive convergents directly from the recurrence relations. The base case is $n=1$, and the induction step follows because the terms containing $a_n$ cancel. The irrationality of $\theta$ ensures that the simple continued fraction is infinite, so the convergents exist for every index under discussion; once a fixed index is chosen, the determinant calculation uses only the recurrence relations and the positivity of the denominators. Once the determinant identity is established, the desired difference formula follows by dividing by the positive product $q_n q_{n-1}$. [/proofplan] [step:Prove the continuant determinant identity by induction] For each integer $n \geq 1$, define the integer \begin{align*} D_n := p_n q_{n-1}-p_{n-1}q_n. \end{align*} We prove that \begin{align*} D_n = (-1)^{n-1} \end{align*} for every $n \geq 1$. For $n=1$, the recurrence gives $p_0=a_0$, $q_0=1$, $p_1=a_1a_0+1$, and $q_1=a_1$. Therefore \begin{align*} D_1 = p_1q_0-p_0q_1 = (a_1a_0+1)\cdot 1-a_0a_1 = 1 = (-1)^0. \end{align*} Now let $n \geq 2$ and assume $D_{n-1}=(-1)^{n-2}$. Using the recurrence relations for $p_n$ and $q_n$, we compute \begin{align*} D_n = p_nq_{n-1}-p_{n-1}q_n. \end{align*} Substituting $p_n=a_np_{n-1}+p_{n-2}$ and $q_n=a_nq_{n-1}+q_{n-2}$ gives \begin{align*} D_n = (a_np_{n-1}+p_{n-2})q_{n-1}-p_{n-1}(a_nq_{n-1}+q_{n-2}). \end{align*} Expanding and cancelling the two identical terms with opposite signs gives \begin{align*} D_n = p_{n-2}q_{n-1}-p_{n-1}q_{n-2}. \end{align*} By the definition of $D_{n-1}$, \begin{align*} D_n = -D_{n-1}. \end{align*} The induction hypothesis now gives \begin{align*} D_n = -(-1)^{n-2}=(-1)^{n-1}. \end{align*} By induction, $D_n=(-1)^{n-1}$ for every $n \geq 1$. [guided] The quantity we need to understand is the numerator obtained after subtracting two consecutive fractions. For that reason, define \begin{align*} D_n := p_n q_{n-1}-p_{n-1}q_n \end{align*} for every $n \geq 1$. This is the determinant of the two consecutive numerator-denominator columns, and it is exactly the numerator that appears when the fractions are put over the common denominator $q_nq_{n-1}$. We first compute the base case $n=1$. In the standard recurrence for continued-fraction convergents, the auxiliary initial values are \begin{align*} p_{-1}=1, \qquad q_{-1}=0, \end{align*} while $p_0=a_0$ and $q_0=1$. Therefore the recurrence gives \begin{align*} p_1=a_1p_0+p_{-1}=a_1a_0+1, \qquad q_1=a_1q_0+q_{-1}=a_1. \end{align*} Hence \begin{align*} D_1 = p_1q_0-p_0q_1 = (a_1a_0+1)\cdot 1-a_0a_1 = 1 = (-1)^0. \end{align*} This establishes the determinant identity at the first nontrivial pair of consecutive convergents. Now let $n \geq 2$ and assume the identity is known at the previous index: \begin{align*} D_{n-1}=p_{n-1}q_{n-2}-p_{n-2}q_{n-1}=(-1)^{n-2}. \end{align*} We compute $D_n$ using only the recurrence relations. Substituting $p_n=a_np_{n-1}+p_{n-2}$ and $q_n=a_nq_{n-1}+q_{n-2}$, we obtain \begin{align*} D_n = (a_np_{n-1}+p_{n-2})q_{n-1}-p_{n-1}(a_nq_{n-1}+q_{n-2}). \end{align*} Expanding the products gives \begin{align*} D_n = a_np_{n-1}q_{n-1}+p_{n-2}q_{n-1}-a_np_{n-1}q_{n-1}-p_{n-1}q_{n-2}. \end{align*} The two terms containing $a_n$ cancel because they are identical with opposite signs. What remains is \begin{align*} D_n = p_{n-2}q_{n-1}-p_{n-1}q_{n-2}. \end{align*} By rewriting the order of the two terms, this is \begin{align*} D_n = -\bigl(p_{n-1}q_{n-2}-p_{n-2}q_{n-1}\bigr) = -D_{n-1}. \end{align*} Using the induction hypothesis gives $D_n=-(-1)^{n-2}=(-1)^{n-1}$. Thus the determinant alternates sign at each step, and the identity \begin{align*} p_nq_{n-1}-p_{n-1}q_n = (-1)^{n-1} \end{align*} holds for every $n \geq 1$. [/guided] [/step] [step:Divide the determinant identity by the positive denominator product] Fix $n \geq 1$. Since $q_n>0$ and $q_{n-1}>0$, the product $q_nq_{n-1}$ is positive and nonzero. Therefore subtraction of fractions gives \begin{align*} \frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}} = \frac{p_nq_{n-1}-p_{n-1}q_n}{q_nq_{n-1}}. \end{align*} Using the determinant identity proved above, \begin{align*} \frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}} = \frac{(-1)^{n-1}}{q_nq_{n-1}}. \end{align*} This is the desired formula for every $n \geq 1$. [/step]