Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 227: 47


$$\tan(\pi-x)=-\tan x$$

Work Step by Step

$$\tan(\pi-x)$$ Apply the identity of tangent of a difference here, we have $$=\frac{\tan\pi-\tan x}{1+\tan\pi\tan x}$$ We know that $\tan\pi=0$. If you do not already know, see that $\tan\pi=\frac{\sin\pi}{\cos\pi}$, in which $\sin\pi$ equals $0$ and $\cos\pi$ equals $1$. That means, $$\tan(\pi-x)=\frac{0-\tan x}{1-0\times \tan x}$$ $$=\frac{-\tan x}{1}$$ $$=-\tan x$$
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