where are the vertical asymptotes of y = tan x between $-pi$ and $pi$? $-\frac{pi}{?}$ and $\frac{pi}{}$

Explanation:

Step1: Recall the definition of tangent

The tangent function is defined as $\tan x = \frac{\sin x}{\cos x}$. Vertical asymptotes occur where the denominator is zero (and the numerator is not zero) because the function is undefined there and approaches $\pm\infty$.

Step2: Find when $\cos x = 0$

The solutions to $\cos x = 0$ are $x = \frac{\pi}{2} + k\pi$, where $k$ is an integer.

Step3: Find solutions in the interval $(-\pi, \pi)$

For $k = -1$: $x = \frac{\pi}{2} - \pi = -\frac{\pi}{2}$ For $k = 0$: $x = \frac{\pi}{2}$ These are the values of $x$ in $(-\pi, \pi)$ where $\cos x = 0$ (and $\sin x \neq 0$), so these are the vertical asymptotes.

Answer:

The vertical asymptotes of $y = \tan x$ between $-\pi$ and $\pi$ are at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$. So the missing denominator in both cases is $2$.