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Q. 98

Expert-verified
Found in: Page 487

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Show that ${\mathrm{tan}}^{-1}v+co{t}^{-1}v=\frac{\mathrm{\pi }}{2}$

The equation ${\mathrm{tan}}^{-1}v+co{t}^{-1}v=\frac{\mathrm{\pi }}{2}$can be obtained as

$x={\mathrm{tan}}^{-1}\left(v\right)\phantom{\rule{0ex}{0ex}}y=\mathrm{co}{t}^{-1}\left(v\right)\phantom{\rule{0ex}{0ex}}so\mathrm{tan}\left(x\right)=v\phantom{\rule{0ex}{0ex}}\mathrm{co}t\left(y\right)=v\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(x\right)=\mathrm{co}t\left(y\right)\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(x\right)=\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-y\right)\phantom{\rule{0ex}{0ex}}x=\frac{\mathrm{\pi }}{2}-y\phantom{\rule{0ex}{0ex}}x+y=\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{-1}\left(v\right)+\mathrm{co}{t}^{-1}\left(v\right)=\frac{\mathrm{\pi }}{2}$

See the step by step solution

## Step 1. Given data

The given equation for verification is

${\mathrm{tan}}^{-1}v+co{t}^{-1}v=\frac{\mathrm{\pi }}{2}$

## Step 2. Considering variables

Consider

$x={\mathrm{tan}}^{-1}\left(v\right)\phantom{\rule{0ex}{0ex}}y=co{t}^{-1}\left(v\right)$

Change inverse trigonometric into normal trigonometric function

$\mathrm{tan}\left(x\right)=v\phantom{\rule{0ex}{0ex}}\mathrm{co}t\left(y\right)=v$

so $\mathrm{tan}\left(x\right)=\mathrm{co}t\left(y\right)$

## Step 3. proof

Use formula

$\mathrm{co}t\left(n\right)=\mathrm{tan}\left(\frac{\mathrm{\pi }}{2}-n\right)$

so$\mathrm{tan}\left(x\right)=\mathrm{co}t\left(y\right)\phantom{\rule{0ex}{0ex}}\mathrm{tan}\left(x\right)=\mathrm{tan}\left(\frac{\mathrm{\pi }}{2}-y\right)\phantom{\rule{0ex}{0ex}}x=\frac{\mathrm{\pi }}{2}-y\phantom{\rule{0ex}{0ex}}x+y=\frac{\mathrm{\pi }}{2}$

Substitute expression of x and y

$x+y=\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{-1}v+co{t}^{-1}v=\frac{\mathrm{\pi }}{2}$