Question

Show that${\tan }^{-1}\left(\frac{1}{v}\right)=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}v$, if$v>0$

Short answer

The equation ${\tan }^{-1}\left(\frac{1}{v}\right)=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}v$can be obtained by the formula

$$\begin{gathered} {\tan }^{-1}v+co{t}^{-1}v=\frac{\mathrm{\pi }}{2} \\ co{t}^{-1}v=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}v \\ {\tan }^{-1}v+co{t}^{-1}v=\frac{\mathrm{\pi }}{2} \\ {\tan }^{-1}\left(\frac{1}{v}\right)=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}v \end{gathered}$$

Step-by-step solution

Step 1. Given data 

The given equation for verification is

${\tan }^{-1}\left(\frac{1}{v}\right)=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}v$

Step 2. Consider

Consider

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

Step 3. proof 

Use formula

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

From equation $1$

$$\begin{gathered} co{t}^{-1}v=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}v \\ {\tan }^{-1}\left(\frac{1}{v}\right)=\frac{\mathrm{\pi }}{2}-{\tan }^{-1}v \end{gathered}$$