Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Cumulative Review Exercises - Page 880: 37


See the explanation given below.

Work Step by Step

Let us consider the left side: $\tan x+\tan y $ We know that, $\tan x=\frac{\sin x}{\cos x}$. It implies: $\begin{align} & \tan x+\tan y=\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y} \\ & =\frac{\sin x\cos y+\sin y\cos x}{\cos x\cos y} \end{align}$ By using the identity $\sin \left( x+y \right)=\sin x\cos y+\sin y\cos x $, we obtain: $\tan x+\tan y=\frac{\sin \left( x+y \right)}{\cos x\cos y}$ Thus, $\tan x+\tan y=\frac{\sin \left( x+y \right)}{\cos x\cos y}$
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