## Precalculus (10th Edition)

Published by Pearson

# Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 395: 88

#### Answer

$0$

#### Work Step by Step

I know that $\cos$ has a period of $360^\circ$, hence $\cos{\theta}=\cos{(\theta+360^\circ)}.$ Thus $\cos{-430^\circ}=\cos{-430^\circ+360^\circ}=\cos{-70^\circ}$. I know that $\cos$ is an even function, which means $f(-\theta)=f(\theta).$ Therefore $\cos{-70^\circ}=\cos{70^\circ}$. I know that $\tan$ is an even function, which means $f(-\theta)=-f(\theta).$ Therefore $\tan{-70^\circ}=-\tan{70^\circ}$. Thus: $\frac{\sin{70^\circ}}{\cos{-430^\circ}}+\tan{-70^\circ}=\frac{\sin{70^\circ}}{\cos{70^\circ}}-\tan{70^\circ}=\tan{70^\circ}-\tan{70^\circ}=0$ Because $\frac{\sin{\theta}}{\cos{\theta}}=\tan{\theta}$.

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