Question

give the degree measure of θ. θ = cot⁻¹(-1) a. 90° b. 180° c. 135° d. 45°

Explanation:

Step1: Recall the definition of arccotangent function

The range of \(y = \cot^{-1}(x)\) is \((0,\pi)\) (or \((0^{\circ},180^{\circ})\) in degree measure). Also, \(\cot\theta=\frac{\cos\theta}{\sin\theta}\), and \(\cot^{-1}(x)=\theta\) means \(\cot\theta = x\).

Step2: Use the identity \(\cot\theta=\frac{1}{\tan\theta}\)

If \(\theta=\cot^{-1}(- 1)\), then \(\cot\theta=-1\), and since \(\cot\theta=\frac{1}{\tan\theta}\), we have \(\tan\theta=-1\).

Step3: Find the angle in the range \((0^{\circ},180^{\circ})\)

We know that \(\tan\theta=-1\). The general solution of \(\tan\theta = k\) is \(\theta=n\pi+\arctan(k)\), \(n\in\mathbb{Z}\). For \(n = 1\) and \(k=-1\), and considering the range \(\theta\in(0^{\circ},180^{\circ})\), \(\tan(135^{\circ})=\tan(180^{\circ} - 45^{\circ})=-\tan(45^{\circ})=-1\)

Answer:

C. \(135^{\circ}\)