3. If $0<x<\frac{\pi}{2}$ and $\sin x-\cos x=\frac{\pi}{4}$ and $\tan x+\frac{1}{\tan x}=\frac{a}{b-\pi^{c}}$, where $a, b$ and $c$ are positive integers, find the value of $a+b+c$.
(A) 8
(B) 32
(C) 34
(D) 48
(E) 50
3. Answer: (E)
We have $(\sin x-\cos x)^{2}=\frac{\pi^{2}}{16}$, which implies that $\sin x \cos x=\frac{16-\pi^{2}}{32}$.
Therefore we obtain
$$
\tan x+\frac{1}{\tan x}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{1}{\sin x \cos x}=\frac{32}{16-\pi^{2}} .
$$

Hence $a+b+c=32+16+2=50$,