Suggested languages for you:

Americas

Europe

Q. 100

Expert-verified
Found in: Page 487

### Precalculus Enhanced with Graphing Utilities

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

# Show that$co{t}^{-1}{e}^{v}={\mathrm{tan}}^{-1}{e}^{-v}$

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

$co{t}^{-1}{e}^{v}=A\cdots \left(i\right)\phantom{\rule{0ex}{0ex}}{e}^{v}=cotA\phantom{\rule{0ex}{0ex}}{e}^{-v}=\mathrm{tan}A\phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{-1}{e}^{v}=A\cdots \left(ii\right)$

from $i&ii$

$co{t}^{-1}{e}^{v}={\mathrm{tan}}^{-1}{e}^{-v}$

See the step by step solution

## Step 1. Given data

The given equation for verification is

$co{t}^{-1}{e}^{v}={\mathrm{tan}}^{-1}{e}^{-v}$

## Step 2. Proof

Consider

$co{t}^{-1}{e}^{v}=A\cdots \left(1\right)$

Change inverse trigonometric into normal trigonometric function

${e}^{v}=cotA$

and$\frac{1}{{e}^{v}}=\frac{1}{cotA}\phantom{\rule{0ex}{0ex}}{e}^{-v}=\mathrm{tan}A\phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{-1}\left({e}^{-v}\right)=A\cdots \left(2\right)$

From Equation $1&2$

$co{t}^{-1}{e}^{v}={\mathrm{tan}}^{-1}{e}^{-v}$