Question

(a) Show that$\tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=0$

(b) Conclude from part (a) that

$\tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=0$

Short answer

(a)The Equation$\tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=0$can be obtained by using the sum formula for tangent function as

$$\begin{gathered} \tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=\frac{\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)+3}{1-\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)·3} \\ =\frac{\frac{\tan \left({\tan }^{-1}1\right)+\tan \left({\tan }^{-1}2\right)}{1-\tan \left({\tan }^{-1}1\right)·\tan \left({\tan }^{-1}2\right)}+3}{1-\frac{\tan \left({\tan }^{-1}1\right)+\tan \left({\tan }^{-1}2\right)}{1-\tan \left({\tan }^{-1}1\right)·\tan \left({\tan }^{-1}2\right)}·3} \\ =0 \end{gathered}$$

(b) the equation ${\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3=\mathrm{\pi }$is obtained as

$$\begin{gathered} \tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=0 \\ {\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3={\tan }^{-1}0 \\ {\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3=\mathrm{\pi } \end{gathered}$$

Step-by-step solution

Step 1. Given data

Equations that need to be proven are
$$\begin{gathered} \tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=0 \\ {\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3=\mathrm{\pi } \end{gathered}$$

Step 2. Part (a)

Use Sum formula for the tangent function

$$\begin{gathered} \tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=\tan \left(\left({\tan }^{-1}1+{\tan }^{-1}2\right)+{\tan }^{-1}3\right) \\ =\frac{\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)+\tan \left({\tan }^{-1}3\right)}{1-\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)·\tan \left({\tan }^{-1}3\right)} \\ =\frac{\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)+3}{1-\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)·3} \end{gathered}$$

Step 3. Apply Sum formula for the tangent function

Use Sum formula for the tangent function

$$\begin{gathered} \tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=\frac{\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)+3}{1-\tan \left({\tan }^{-1}1+{\tan }^{-1}2\right)·3} \\ =\frac{\frac{\tan \left({\tan }^{-1}1\right)+\tan \left({\tan }^{-1}2\right)}{1-\tan \left({\tan }^{-1}1\right)·\tan \left({\tan }^{-1}2\right)}+3}{1-\frac{\tan \left({\tan }^{-1}1\right)+\tan \left({\tan }^{-1}2\right)}{1-\tan \left({\tan }^{-1}1\right)·\tan \left({\tan }^{-1}2\right)}·3} \\ =\frac{\frac{1+2}{1-1\times 2}+3}{1-\frac{1+2}{1-1\times 2}·3} \\ =\frac{-3+3}{1+9} \\ =0 \end{gathered}$$

so$\tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=0$

Step 4. Part (b)

Change inverse trigonometric function into a trigonometric function

$$\begin{gathered} \tan \left({\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3\right)=0 \\ {\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3={\tan }^{-1}0 \end{gathered}$$

for interval$[0,\mathrm{\pi }]$

$$\begin{gathered} {\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3={\tan }^{-1}0 \\ {\tan }^{-1}1+{\tan }^{-1}2+{\tan }^{-1}3=\mathrm{\pi } \end{gathered}$$