Chapter 4.3, Problem 49E

To determine

To construct: ${\tan }^{-1}1,{\tan }^{-1}2\text{and}{\tan }^{-1}3$ by folding a square of paper. Try it and explain what is going on.

Answer to Problem 49E

  ${\tan }^{-1}(1)+{\tan }^{-1}(2)+{\tan }^{-1}(3)=\pi .$

Explanation of Solution

Given information: Here is a way to construct ${\tan }^{-1}1,{\tan }^{-1}2\text{and}{\tan }^{-1}3$ by folding a

square of paper as shown in the below figure.

  

Calculation:

From the given figure need to prove that,

  $\begin{array}{c}{\tan }^{-1}(1)+{\tan }^{-1}(2)+{\tan }^{-1}(3)=\pi .\\ {\tan }^{-1}(1)+{\tan }^{-1}(2)=\pi -{\tan }^{-1}(3).\\ \text{Taking}\text{tan}\text{of}\text{both}\text{sided}.\\ \tan ({\tan }^{-1}(1)+{\tan }^{-1}(2))=\tan (\pi -{\tan }^{-1}(3)).\\ \tan (\pi -{\tan }^{-1}(3))=-\tan ({\tan }^{-1}(3))(\text{since}\tan (\pi -x)=-\tan x).\\ \tan (\pi -{\tan }^{-1}(3))=-3(\text{since}\tan ({\tan }^{-1}(x))=x\text{for}\text{all}x\text{in}\text{real}\text{number}s).\\ \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A.\tan B}.\\ \text{Therefore},\\ \tan ({\tan }^{-1}(1)+{\tan }^{-1}(2))=\frac{\tan ({\tan }^{-1}(1))+\tan ({\tan }^{-1}(2))}{1-\tan ({\tan }^{-1}(1))\tan ({\tan }^{-1}(2))}.\\ \tan ({\tan }^{-1}(1)+{\tan }^{-1}(2))=\frac{1+2}{1-1(2)}=\frac{3}{-1}=-3.\\ \text{So,}\\ -3=-3.\end{array}$

Hence proved.