Question

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

Short answer

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

$$\begin{gathered} co{t}^{-1}{e}^v=A\cdots \left(i\right) \\ {e}^v=cotA \\ {e}^{-v}=\tan A \\ {\tan }^{-1}{e}^v=A\cdots \left(ii\right) \end{gathered}$$

from $i\&ii$

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

Step-by-step solution

Step 1. Given data 

The given equation for verification is

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

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$$\begin{gathered} \frac{1}{e^v}=\frac{1}{cotA} \\ {e}^{-v}=\tan A \\ {\tan }^{-1}\left(e^{-v}\right)=A\cdots \left(2\right) \end{gathered}$$

From Equation $1\&2$

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