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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 98 (page 487)

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

Short Answer

Expert verified

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

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

Step by step solution

01

Step 1. Given data

The given equation for verification is

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

02

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

03

Step 3. proof

Use formula

$co t (n) = \tan (\frac{蟺}{2} - n)$

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

Substitute expression of xand y

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

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