Chapter 5.5, Problem 47E

(a)

To determine

To graph:The function and make a conjecture

Explanation of Solution

Given: The given function is $y={\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}$ .

The graph of the function $y={\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}$ is shown in figure below.

  

Figure (1)

From the figure the conjecture is,

  ${\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=\{\begin{array}{l}\frac{\pi }{2}x>0\\ \frac{-\pi }{2}x<0\end{array}$

Therefore, the graph of the function is shown in figure (1) and its conjecture is ${\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=\{\begin{array}{l}\frac{\pi }{2}x>0\\ \frac{-\pi }{2}x<0\end{array}$ .

(b)

To determine

To prove:The conjecture ${\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=\{\begin{array}{l}\frac{\pi }{2}x>0\\ \frac{-\pi }{2}x<0\end{array}$

Explanation of Solution

Let ${\tan }^{-1}x=t$

So,

  $\begin{array}{c}\tan t=x\\ \frac{1}{\tan t}=\frac{1}{x}\\ \cot t=\frac{1}{x}\\ \tan \left(\frac{\pi }{2}-t\right)=\frac{1}{x},x>0\end{array}$

Solve further,

  $\begin{array}{c}\frac{\pi }{2}-t={\tan }^{-1}\left(\frac{1}{x}\right)\\ {\tan }^{-1}x+{\tan }^{-1}\left(\frac{1}{x}\right)=\frac{\pi }{2}\end{array}$

Also,

  $\begin{array}{c}\tan \left(-\frac{\pi }{2}-t\right)=\frac{1}{x}\\ \frac{-\pi }{2}-t={\tan }^{-1}\left(\frac{1}{x}\right)\\ {\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=\frac{-\pi }{2},x<0\end{array}$

Hence, proved.