Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 14 - Trigonometric Identities and Equations - 14-1 Trigonometric Identities - Lesson Check - Page 908: 3

Answer

$\sec\theta-\cos\theta=\sec\theta-\cos\theta$

Work Step by Step

Work on the left hand side: Step 1: Replace $\tan\theta$ with $\frac{\sin\theta}{\cos\theta}$ by definition. Hence, $\sin\theta\times \tan\theta= \sin\theta\times\frac{\sin\theta}{cos\theta}$ $\sin\theta\times \tan\theta= \frac{\sin^{2}\theta}{\cos\theta}$ Step 2: By identity, $\sin^{2}\theta+\cos^{2}\theta=1$. Replace $\sin^{2}\theta$ by $1-\cos^{2}\theta$ Hence, $\sin\theta\times \tan\theta = \frac{1-\cos^{2}\theta}{\cos\theta}$ Step 3: Split the denominator across two terms $\sin\theta\times \tan\theta$= $\frac{1}{\cos\theta} - \frac{\cos^{2}\theta}{\cos\theta}$ Step 4: Cancel 1 $cos\theta$ in the 2nd term and replace the $\frac{1}{\cos\theta}$ in the first term by $\sec\theta$ by definition Hence, $\sin\theta\times \tan\theta = \sec\theta-\cos\theta$
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