Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.2 Verifying Trigonometric Identities - 5.2 Exercises - Page 202: 46

Answer

$$\sec^4 x-\sec^2 x=\tan^4 x+\tan^2 x$$ The equation is verified to be an identity.

Work Step by Step

$$\sec^4 x-\sec^2 x=\tan^4 x+\tan^2 x$$ 2 sides look equally complex, so any choice of simplification is arbitrary. Here I would deal with the left side first, using $\sec x=\frac{1}{\cos x}$ $$\sec^4 x-\sec^2 x$$ $$=\frac{1}{\cos^4 x}-\frac{1}{\cos^2 x}$$ $$=\frac{1-\cos^2 x}{\cos^4 x}$$ As we already know, $\sin^2 x+\cos^2 x=1$. But that also means $1-\cos^2 x=\sin^2 x$. Therefore, $$=\frac{\sin^2 x}{\cos^4 x}$$ On the right side, again we simplify, using $\tan x=\frac{\sin x}{\cos x}$ $$\tan^4 x+\tan^2 x$$ $$=\frac{\sin^4 x}{\cos^4 x}+\frac{\sin^2 x}{\cos^2 x}$$ $$=\frac{\sin^4 x}{\cos^4 x}+\frac{\sin^2 x\cos^2 x}{\cos^4 x}$$ $$=\frac{\sin^4 x+\sin^2 x\cos^2 x}{\cos^4 x}$$ $$=\frac{\sin^2 x(\sin^2 x+\cos^2 x)}{\cos^4 x}$$ $$=\frac{\sin^2 x}{\cos^4 x}$$ (for $\sin^2 x+\cos^2 x=1$) Both sides equal $\frac{\sin^2 x}{\cos^4 x}$, so they are equal. $$\sec^4 x-\sec^2 x=\tan^4 x+\tan^2 x$$ The equation is thus verified to be an identity.
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