Question

use reference angles to evaluate the expression.
(\tan left(225^{circ}
ight)=)
question help: (square) message instructor
question 5
use a graphing calculator to evaluate. round to 3 decimal places.
(sec left(162^{circ}
ight)=)

Explanation:

Step1: Find the reference angle for \(225^{\circ}\)

Since \(225^{\circ}=180^{\circ} + 45^{\circ}\), the reference angle \(\theta'=225^{\circ}-180^{\circ}=45^{\circ}\). And \(225^{\circ}\) is in the third - quadrant. In the third - quadrant, \(\tan\theta=\frac{\sin\theta}{\cos\theta}\), and both \(\sin\theta\) and \(\cos\theta\) are negative, so \(\tan\theta>0\). Then \(\tan(225^{\circ})=\tan(45^{\circ})\).

Step2: Evaluate \(\tan(45^{\circ})\)

We know that \(\tan(45^{\circ}) = 1\).

Step3: Evaluate \(\sec(162^{\circ})\)

First, use the identity \(\sec\theta=\frac{1}{\cos\theta}\). So \(\sec(162^{\circ})=\frac{1}{\cos(162^{\circ})}\).
The reference angle for \(162^{\circ}\) is \(\theta'=180^{\circ}-162^{\circ} = 18^{\circ}\). And \(162^{\circ}\) is in the second - quadrant, where \(\cos\theta<0\). So \(\cos(162^{\circ})=-\cos(18^{\circ})\).
Using a calculator, \(\cos(18^{\circ})\approx0.951\), then \(\cos(162^{\circ})\approx - 0.951\), and \(\sec(162^{\circ})=\frac{1}{\cos(162^{\circ})}\approx\frac{1}{- 0.951}\approx - 1.052\)

Answer:

\(\tan(225^{\circ}) = 1\); \(\sec(162^{\circ})\approx - 1.052\)