Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Practice Exercises - Page 636: 56



Work Step by Step

Consider the Maclaurin Series for $ \tan^{-1} x$ as follows: $ \tan^{-1} x=\Sigma_{n=0}^\infty (-1)^n\dfrac{x^{2n-1}}{(2n-1)}=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+...$ As we are given that $x=\dfrac{1}{\sqrt 3}$ Now, we have $\tan^{-1} x= \tan^{-1} (\dfrac{1}{\sqrt 3})$ $\implies \tan^{-1} (\tan (\dfrac{\pi}{6}))=\dfrac{\pi}{6}$
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