## Precalculus (6th Edition)

Published by Pearson

# Chapter 7 - Trigonometric Identities and Equations - 7.3 Sum and Difference Identities - 7.3 Exercises - Page 680: 100

#### Answer

$\frac{\sin(x+y)}{\cos(x-y)}=\frac{\cot x+\cot y}{1+\cot x\cot y}$

#### Work Step by Step

Start with the right side: $\frac{\cot x+\cot y}{1+\cot x\cot y}$ Express it in terms of sine and cosine: $=\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1+\frac{\cos x}{\sin x}*\frac{\cos y}{\sin y}}$ Multiply top and bottom by $\sin x\sin y$: $=\frac{\frac{\cos x}{\sin x}+\frac{\cos y}{\sin y}}{1+\frac{\cos x}{\sin x}*\frac{\cos y}{\sin y}}*\frac{\sin x\sin y}{\sin x\sin y}$ $=\frac{\cos x\sin y+\sin x\cos y}{\sin x\sin y+\cos x\cos y}$ $=\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y+\sin x\sin y}$ Use sum and difference identities for sine and cosine to simplify: $=\frac{\sin(x+y)}{\cos(x-y)}$ Since this equals the left side, the identity has been proven.

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