Determine the intervals of monotonicity of the function y

Determine the intervals of monotonicity of the function $\displaystyle y = \frac{\sin x-2}{\cos x}$.

Step 1: Domain

The function is defined wherever $\cos x\neq0$, that is on the union of the intervals

\[ I_{k} = \left(-\frac{\pi}{2}+k\pi,\;\frac{\pi}{2}+k\pi\right), \qquad k\in\mathbb{Z}. \]

Every statement about monotonicity refers to one such interval at a time: across a pole nothing can be compared.

Step 2: The derivative

By the quotient rule with $u = \sin x-2$, $v = \cos x$: \[y' = \frac{\cos x\cdot\cos x-\bigl(\sin x-2\bigr)\bigl(-\sin x\bigr)}{\cos^{2}x} = \frac{\cos^{2}x+\sin^{2}x-2\sin x}{\cos^{2}x} = \frac{1-2\sin x}{\cos^{2}x}. \tag{1}\]

Step 3: Sign of the derivative

The denominator $\cos^{2}x$ is positive throughout the domain, so the sign of $y'$ is the sign of $1-2\sin x$; it vanishes where $\sin x = \dfrac12$, i.e. at $x = \dfrac{\pi}{6}+2k\pi$ and $x = \dfrac{5\pi}{6}+2k\pi$. Exactly one of these lies in each interval $I_{k}$.

Step 4: Intervals containing $\dfrac{\pi}{6}$ modulo $2\pi$

On $\left(-\dfrac{\pi}{2}+2m\pi,\;\dfrac{\pi}{2}+2m\pi\right)$ the sine increases from $-1$ to $1$ and passes $\tfrac12$ at $x = \dfrac{\pi}{6}+2m\pi$: \[\left(-\frac{\pi}{2}+2m\pi,\;\frac{\pi}{6}+2m\pi\right):\ \sin x\lt \tfrac12 \Rightarrow y'\gt 0\ (\nearrow); \qquad \left(\frac{\pi}{6}+2m\pi,\;\frac{\pi}{2}+2m\pi\right):\ y'\lt 0\ (\searrow).\]

Step 5: Intervals containing $\dfrac{5\pi}{6}$ modulo $2\pi$

On $\left(\dfrac{\pi}{2}+2m\pi,\;\dfrac{3\pi}{2}+2m\pi\right)$ the sine decreases from $1$ to $-1$ and passes $\tfrac12$ at $x = \dfrac{5\pi}{6}+2m\pi$: \[\left(\frac{\pi}{2}+2m\pi,\;\frac{5\pi}{6}+2m\pi\right):\ \sin x\gt \tfrac12 \Rightarrow y'\lt 0\ (\searrow); \qquad \left(\frac{5\pi}{6}+2m\pi,\;\frac{3\pi}{2}+2m\pi\right):\ y'\gt 0\ (\nearrow).\]

Step 6: Verification

\[y\!\left(-\tfrac{\pi}{3}\right) = -5.7320508 \lt y(0) = -2 \lt y\!\left(\tfrac{\pi}{6}\right) = -1.7320508 \;\;\checkmark\ (\nearrow)\]

\[y\!\left(\tfrac{5\pi}{6}\right) = 1.7320508 \lt y\bigl(\pi\bigr) = 2 \;\;\checkmark\ (\nearrow \text{ after the turning point})\]

\[\text{Local maximum } -\sqrt3 \text{ at } \tfrac{\pi}{6}+2k\pi;\ \text{local minimum } \sqrt3 \text{ at } \tfrac{5\pi}{6}+2k\pi \;\;\checkmark\]