Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 6 - Section 6.4 - Inverse Trigonometric Functions and Right Triangles - 6.4 Exercises - Page 507: 32

Answer

$\frac{25}{24}$

Work Step by Step

Given expression is- $\csc(\cos^{-1}\frac{7}{25})$ Assuming $\cos^{-1} \frac{7}{25}$ = $u$ , we get- $\cos u$ = $\frac{7}{25}$, [$0\leq u\leq \pi$] From First Pythagorean identity- $\sin u$ = + $\sqrt {1 - \cos^{2} u}$ = $\sqrt {1 - (\frac{7}{25})^{2}} $ = $\sqrt {1 - \frac{49}{625}} $ = $\sqrt { \frac{625 - 49}{625}} $ = $\sqrt { \frac{576}{625}} $ = $\frac{24}{25}$ i.e. $\sin u$ = $\frac{24}{25}$ Therefore $\csc u$ = $\frac{1}{\sin u}$ = $\frac{1}{\frac{24}{25}}$ i.e. $\csc u$ = $\frac{25}{24}$ i.e. $\csc(\cos^{-1}\frac{7}{25})$ = $\frac{25}{24}$
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