Algebra 2 (1st Edition)

Published by McDougal Littell

Chapter 13, Trigonometric Ratios and Functions - 13.4 Evaluate Inverse Trigonometric Functions - 13.4 Exercises - Quiz for Lessons 13.3-13.4 - Page 880: 1

Answer

See answer below.

Work Step by Step

The Trigonometric Identities are written as below: $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{-2}{2 \sqrt {10}}=-\dfrac{\sqrt {10}}{\sqrt {10}}$ $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{6}{2 \sqrt {10}}=\dfrac{3 \sqrt {10}}{10}$ $\tan \theta=\dfrac{Opposite}{Adjacent}=\dfrac{-2}{6}=-\dfrac{1}{3}$ $\csc \theta=\dfrac{Hypotenuse}{Opposite}=\dfrac{6}{-2}=-3$ $\sec \theta=\dfrac{Hypotenuse}{Adjacent}=\dfrac{2 \sqrt {10}}{6}=\dfrac{\sqrt {10}}{3}$ $\cot \theta=\dfrac{Adjacent}{Opposite}=\dfrac{2 \sqrt {10}}{-2}=-\sqrt{10}$

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