1. Given $0<x<\frac{\pi}{2}, \sin x-\cos x=\frac{\pi}{4}$, if $\tan x+\frac{1}{\tan x}$ can be expressed in the form $\frac{a}{b-\pi^{c}}$, where $a$, $b$, and $c$ are positive integers, then $a+b+c=(\quad)$.
(A) 8
(B) 32
(C) 48
(D) 50
(1) From the given condition, $(\sin x-\cos x)^{2}=\left(\frac{\pi}{4}\right)^{2}$, therefore $\sin x \cos x=\frac{16-\pi^{2}}{32}$, so
$$
\begin{aligned}
\tan x+\frac{1}{\tan x} & =\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \\
& =\frac{1}{\sin x \cos x} \\
& =\frac{32}{16-\pi^{2}} .
\end{aligned}
$$

Thus $a+b+c=32+16+2=50$. Therefore, the answer is D.