## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 9 - 9.2 - Two-Variable Linear Systems - 9.2 Exercises - Page 649: 70

#### Answer

$u=-\dfrac{\sin (2x) \csc x}{2}$ and $v=\dfrac{\cos (2x) \csc x}{2}$

#### Work Step by Step

Rewrite the first equation as: $u=\dfrac{-v \sin 2x}{\cos 2x}$ Substitute the value of $u$ in the first equation to get the value of $v$. $2v \times \dfrac{\sin^2 2x+\cos^2 2x}{\cos (2x)}=\sec x$ Therefore, the above equation yields: $\dfrac{2v}{\cos x}=\sec x \implies v =\dfrac{\cos (2x) \csc x}{2}$ Substitute the value of $v$ in the first equation to get the value of $u$. $u=\dfrac{-(\dfrac{\cos (2x) \csc x}{2}) \times \sin 2x}{\cos 2x}=-\dfrac{\sin (2x) \csc x}{2}$ So, the solution is $u=-\dfrac{\sin (2x) \csc x}{2}$ and $v=\dfrac{\cos (2x) \csc x}{2}$

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