Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 5 - Section 5.1 - Proving Identities - 5.1 Problem Set - Page 280: 68

Answer

The identity is true. For proof, please see "step-by-step".

Work Step by Step

Recognize a difference of cubes in the numerator. $x^{3}-y^{y}=(x-y)(x^{2}+xy+y^{2})$ $LHS=\displaystyle \frac{1-\tan^{3}t}{1-\tan t}=$ ... substituting $x=1, y=\tan t$ $\displaystyle \frac{x^{3}-y^{3}}{x-y}=\frac{(x-y)(x^{2}+xy+y^{2})}{(x-y)}$ ... reduce $=x^{2}+xy+y^{2}$ ...back substitute $=1^{2}+\tan t+\tan^{2}t$ ... apply Pyth. identity: $1 +\tan^{2}\theta=\sec^{2}\theta$ $= \sec^{2}\theta+\tan t=RHS$, proving the identity.
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