Let 0 be an angle in quadrant II such that cos0: 51 Find the exact values of csc 0 and tan0. csce tan 0 %3D

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Let 0 be an angle in quadrant II such that cos0=-2/5
**Trigonometric Values in Quadrant II**

Given:
Let \(\theta\) be an angle in quadrant II such that \( \cos \theta = -\frac{2}{5} \).

**Objective:**
Find the exact values of \(\csc \theta\) and \(\tan \theta\).

---

### Solution Steps:

**1. Identifying \(\sin \theta\) using Pythagorean Identity:**

Since \(\cos \theta = -\frac{2}{5}\), use the Pythagorean identity:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
\[
\sin^2 \theta + \left( -\frac{2}{5} \right)^2 = 1
\]
\[
\sin^2 \theta + \frac{4}{25} = 1
\]
\[
\sin^2 \theta = 1 - \frac{4}{25} = \frac{21}{25}
\]
\[
\sin \theta = \pm \sqrt{\frac{21}{25}} = \pm \frac{\sqrt{21}}{5}
\]

In quadrant II, \(\sin \theta\) is positive, so:
\[
\sin \theta = \frac{\sqrt{21}}{5}
\]

**2. Finding \(\csc \theta\):**

\[
\csc \theta = \frac{1}{\sin \theta} = \frac{5}{\sqrt{21}}
\]

**3. Finding \(\tan \theta\):**

Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):
\[
\tan \theta = \frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}} = -\frac{\sqrt{21}}{2}
\]

---

**Result:**

- \(\csc \theta = \frac{5}{\sqrt{21}}\)
- \(\tan \theta = -\frac{\sqrt{21}}{2}\)

These values were determined for \(\theta\) in quadrant II where \(\cos \theta = -\frac{2}{5}\), and confirm the relationships of \(\sin \theta\) and \(\tan \theta\) consistent with quadrant
Transcribed Image Text:**Trigonometric Values in Quadrant II** Given: Let \(\theta\) be an angle in quadrant II such that \( \cos \theta = -\frac{2}{5} \). **Objective:** Find the exact values of \(\csc \theta\) and \(\tan \theta\). --- ### Solution Steps: **1. Identifying \(\sin \theta\) using Pythagorean Identity:** Since \(\cos \theta = -\frac{2}{5}\), use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \sin^2 \theta + \left( -\frac{2}{5} \right)^2 = 1 \] \[ \sin^2 \theta + \frac{4}{25} = 1 \] \[ \sin^2 \theta = 1 - \frac{4}{25} = \frac{21}{25} \] \[ \sin \theta = \pm \sqrt{\frac{21}{25}} = \pm \frac{\sqrt{21}}{5} \] In quadrant II, \(\sin \theta\) is positive, so: \[ \sin \theta = \frac{\sqrt{21}}{5} \] **2. Finding \(\csc \theta\):** \[ \csc \theta = \frac{1}{\sin \theta} = \frac{5}{\sqrt{21}} \] **3. Finding \(\tan \theta\):** Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\): \[ \tan \theta = \frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}} = -\frac{\sqrt{21}}{2} \] --- **Result:** - \(\csc \theta = \frac{5}{\sqrt{21}}\) - \(\tan \theta = -\frac{\sqrt{21}}{2}\) These values were determined for \(\theta\) in quadrant II where \(\cos \theta = -\frac{2}{5}\), and confirm the relationships of \(\sin \theta\) and \(\tan \theta\) consistent with quadrant
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