Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 6 - Section 6.3 - Trigonometric Functions of Angles - 6.3 Exercises - Page 501: 76

Answer

See explanations below.

Work Step by Step

We are given $sin^2\theta+cos^2\theta=1$ (a) Divide both sides by $cos^2\theta$, we have $\frac{sin^2\theta}{cos^2\theta}+1=\frac{1}{cos^2\theta}$ which gives $tan^2\theta+1=sec^2\theta$ (b) Divide both sides by $sin^2\theta$, we have $1+\frac{cos^2\theta}{sin^2\theta}=\frac{1}{sin^2\theta}$ which gives $1+cot^2\theta=csc^2\theta$
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