Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 48: 76


$\tan(\cos^{-1}(x)) = \frac{\sqrt{1-x^2}}{x}$

Work Step by Step

We are looking to find $\tan(\cos^{-1}(x))$. We can construct a right triangle that can help us find the value of this expression. Since there is a $\cos^{-1}(x)$ term, we can construct a right triangle with a hypotenuse of $1$, a leg of length $x$, and the final leg with a length of $\sqrt{1-x^2}$. Let's call the angle between the leg with length $x$ and hypotenuse, $\theta$. That means $\cos(\theta) = x$, so $\theta = \cos^{-1}(x)$. This is part of what we are looking for, so we can substitute it back into the expression we are interested in: $\tan(\cos^{-1}(x)) = \tan(\theta) = \frac{\sqrt{1-x^2}}{x}$.
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