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 $\frac{a}{b-\pi^{2}}$ where $(a, b, c$ are positive integers). Then $a+b+c=(\quad)$.
(A) 8
(B) 32
(C) 48
(D) 50
-、1. D.
From the given condition,
$(\sin x-\cos x)^{2}=\left(\frac{\pi}{4}\right)^{2}$.
Therefore, $\sin x \cdot \cos x=\frac{16-\pi^{2}}{32}$.
Hence, $\tan x+\frac{1}{\tan x}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$ $=\frac{1}{\sin x \cdot \cos x}=\frac{32}{16-\pi^{2}}$.
Thus, $a+b+c=32+16+2=50$.