Book edition 6th

Author(s) Sullivan

Pages 1200 pages

ISBN 9780321795465

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Short Answer

Show that $\tan^{- 1} v + c o t^{- 1} v = \frac{π}{2}$

The equation $\tan^{- 1} v + c o t^{- 1} v = \frac{π}{2}$ can be obtained as

$x = \tan^{- 1} (v) y = co t^{- 1} (v) s o \tan (x) = v co t (y) = v \tan (x) = co t (y) \tan (x) = \sin (\frac{π}{2} - y) x = \frac{π}{2} - y x + y = \frac{π}{2} \tan^{- 1} (v) + co t^{- 1} (v) = \frac{π}{2}$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given data

The given equation for verification is

$\tan^{- 1} v + c o t^{- 1} v = \frac{π}{2}$

Step 2. Considering variables

Consider

$x = \tan^{- 1} (v) y = c o t^{- 1} (v)$

Change inverse trigonometric into normal trigonometric function

$\tan (x) = v co t (y) = v$

so $\tan (x) = co t (y)$

Step 3. proof

Use formula

$co t (n) = \tan (\frac{π}{2} - n)$

so $\tan (x) = co t (y) \tan (x) = \tan (\frac{π}{2} - y) x = \frac{π}{2} - y x + y = \frac{π}{2}$

Substitute expression of x and y

$x + y = \frac{π}{2} \tan^{- 1} v + c o t^{- 1} v = \frac{π}{2}$

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Precalculus Enhanced with Graphing Utilities

Answers without the blur.

Short Answer

Show that $\tan^{- 1} v + c o t^{- 1} v = \frac{π}{2}$

Step by Step Solution

Step 1. Given data

Step 2. Considering variables

Step 3. proof

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Show that $\tan^{- 1} v + c o t^{- 1} v = \frac{π}{2}$