25~26高三上·湖北12月多校联考·第8题

C

令$f(x) = \tan x – x\left(0 < x < \frac{\pi}{2}\right)$，

则$f'(x) = \frac{1}{\cos^2 x} – 1 = \frac{1 – \cos^2 x}{\cos^2 x} = \tan^2 x > 0$，
$\therefore f(x)$在$\left(0, \frac{\pi}{2}\right)$上单调递增，$\therefore f\left(\frac{1}{4}\right) > f(0) = 0$，
即$\tan\frac{1}{4} > \frac{1}{4}$，$\therefore \frac{\sin\frac{1}{4}}{\cos\frac{1}{4}} > \frac{1}{4}$，

又$\cos\frac{1}{4} > 0$，$\therefore \sin\frac{1}{4} > \frac{1}{4}\cos\frac{1}{4}$，即$a < b$；

令$g(x) = \sin x + \ln(1 – x)\left(0 < x < 1\right)$，则$g'(x) = \cos x – \frac{1}{1 – x}$，
令$h(x) = g'(x)$，则$h'(x) = -\sin x – \frac{1}{(1 – x)^2} < 0$，
$\therefore g'(x)$在$(0, 1)$上单调递减，
$\therefore g'(x) < g'(0) = 0$，$\therefore g(x)$在$(0, 1)$上单调递减，
$\therefore g\left(\frac{1}{4}\right) < g(0) = 0$，即$\sin\frac{1}{4} < -\ln\frac{3}{4} = \ln\frac{4}{3}$，
$\therefore b < c$.

综上所述：$a < b < c$.

故选：C.