知 $-\\frac{\\pi}{2}<x<0, \\sin x+\\cos x=\\frac{1}{5}$, 求下列各式的值.

(1) $\\sin x-\\cos x$;

(2) $\\frac{1}{\\cos ^{2} x-\\sin ^{2} x}$.
(1) $\\because \\sin x+\\cos x=\\frac{1}{5}$,

$\\therefore(\\sin x+\\cos x)^{2}=\\left(\\frac{1}{5}\\right)^{2}$, 即 $1+2 \\sin x \\cos x=\\frac{1}{25}$,

$\\therefore 2 \\sin x \\cos x=-\\frac{24}{25}$.

$\\because(\\sin x-\\cos x)^{2}=\\sin ^{2} x-2 \\sin x \\cos x+\\cos ^{2} x=1-2 \\sin x \\cos x=1+\\frac{24}{25}=\\frac{49}{25}$,

又 $-\\frac{\\pi}{2}<x<0, \\therefore \\sin x<0, \\quad \\cos x>0$,

$\\therefore \\sin x-\\cos x<0$,

$\\therefore \\sin x-\\cos x=-\\frac{7}{5}$.

(2)由已知条件及(1), 可知 $\\left\{\\begin{array}{l}\\sin x+\\cos x=\\frac{1}{5} \\ \\sin x-\\cos x=-\\frac{7}{5}\\end{array}\\right.$,

解得 $left\\{\\begin{array}{l}\\sin x=-\\frac{3}{5} \\\\ \\cos x=\\frac{4}{5}\\end{array}\\right.$,

$\\therefore \\frac{1}{\\cos ^{2} x-\\sin ^{2} x}=\\frac{1}{\\frac{16}{25}-\\frac{9}{25}}=\\frac{25}{7}$.