Question

find the values of $\theta$ if $0 \leq \theta \leq \pi$.
$(\cot \theta + 1)(\sec \theta - \frac{1}{2}) = 0$
$\theta = \frac{?\pi}{\square}$

Explanation:

Step1: Apply Zero-Product Property

Set each factor equal to zero:

1. \(\cot\theta + 1 = 0\)
2. \(\sec\theta - \frac{1}{2} = 0\)

Step2: Solve \(\cot\theta + 1 = 0\)

\(\cot\theta = -1\)
Since \(\cot\theta=\frac{\cos\theta}{\sin\theta}\), we know that \(\cot\theta = -1\) when \(\theta=\frac{3\pi}{4}\) (because in the interval \(0\leq\theta\leq\pi\), \(\cos\frac{3\pi}{4}=-\frac{\sqrt{2}}{2}\), \(\sin\frac{3\pi}{4}=\frac{\sqrt{2}}{2}\), so \(\cot\frac{3\pi}{4}=-1\)).

Step3: Solve \(\sec\theta - \frac{1}{2} = 0\)

\(\sec\theta=\frac{1}{2}\), and \(\sec\theta=\frac{1}{\cos\theta}\), so \(\frac{1}{\cos\theta}=\frac{1}{2}\) implies \(\cos\theta = 2\). But the range of \(\cos\theta\) is \([-1,1]\), so there is no solution for this factor in the interval \(0\leq\theta\leq\pi\).

Answer:

\(\theta = \frac{3\pi}{4}\), so the numerator is \(3\) and the denominator is \(4\).