Trigonometry (11th Edition) Clone

Published by Pearson

Chapter 5 - Review Exercises - Page 248: 7

Answer

$$\sec^2\theta-\tan^2\theta=1$$

Work Step by Step

$$A=\sec^2\theta-\tan^2\theta$$ We need to use 2 identities here, which are $$\sec\theta=\frac{1}{\cos\theta}\hspace{2cm}\tan\theta=\frac{\sin\theta}{\cos\theta}$$ Therefore, $A$ would be $$A=\frac{1}{\cos^2\theta}-\frac{\sin^2\theta}{\cos^2\theta}$$ $$A=\frac{1-\sin^2\theta}{\cos^2\theta}$$ Recall that $1-\sin^2\theta=\cos^2\theta$ from Pythagorean identities. $$A=\frac{\cos^2\theta}{\cos^2\theta}$$ $$A=1$$

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