SOLUTION : $f(x)=\sqrt{\frac{1}{(|x|-1) \cos ^{-1}(2 x+1) \tan 3 x}}$
here $-1 \leq 2 x+1<1 \quad \Rightarrow \quad-2 \leq 2 x<0 \quad \Rightarrow \quad-1 \leq x<0 \quad \Rightarrow \quad x \in[-1,0)$
But $x \neq-1 \quad$ as $\quad|x|-1 \neq 0$
$\therefore \quad x \in(-1,0)$
for $x \in(-1,0),(|x|-1)$ is $-v e$
$\therefore \quad \tan 3 x<0 ; \quad 0>3 x>-\frac{\pi}{2}$ or $\quad x \in\left(-\frac{\pi}{6}, 0\right)$
Domain : $\left(-\frac{\pi}{6}, 0\right) \cap(-1,0) \equiv\left(-\frac{\pi}{6}, 0\right)$
