Book edition 6th

Author(s) Sullivan

Pages 1200 pages

ISBN 9780321795465

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Short Answer

Show that $c o t^{- 1} e^{v} = \tan^{- 1} e^{- v}$

The equation $c o t^{- 1} e^{v} = \tan^{- 1} e^{- v}$ can be obtained as

$c o t^{- 1} e^{v} = A \dots (i) e^{v} = c o t A e^{- v} = \tan A \tan^{- 1} e^{v} = A \dots (i i)$

from $i & i i$

$c o t^{- 1} e^{v} = \tan^{- 1} e^{- v}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given data

The given equation for verification is

$c o t^{- 1} e^{v} = \tan^{- 1} e^{- v}$

Step 2. Proof

Consider

$c o t^{- 1} e^{v} = A \dots (1)$

Change inverse trigonometric into normal trigonometric function

$e^{v} = c o t A$

and $\frac{1}{e^{v}} = \frac{1}{c o t A} e^{- v} = \tan A \tan^{- 1} (e^{- v}) = A \dots (2)$

From Equation $1 & 2$

$c o t^{- 1} e^{v} = \tan^{- 1} e^{- v}$

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Precalculus Enhanced with Graphing Utilities

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Short Answer

Show that $c o t^{- 1} e^{v} = \tan^{- 1} e^{- v}$

Step by Step Solution

Step 1. Given data

Step 2. Proof

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Show that $c o t^{- 1} e^{v} = \tan^{- 1} e^{- v}$