Chapter 14.8, Problem 3E

To determine

The proof of the formula for ${\tan }^{-1}a+{\tan }^{-1}b$ .

Explanation of Solution

To proof the formula,

Let,

  $\begin{array}{l}\tan x=a\\ \tan y=b\end{array}$

Now we know that,

  $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

Now,

  $\begin{array}{l}{\tan }^{-1}(\tan (x+y))={\tan }^{-1}\left(\frac{\tan x+\tan y}{1-\tan x\tan y}\right)\\ x+y={\tan }^{-1}\left(\frac{\tan x+\tan y}{1-\tan x\tan y}\right)\\ \text{since,}\\ \tan x=a\\ \tan y=b\\ Therefore,\\ x={\tan }^{-1}a\\ y={\tan }^{-1}b\\ \text{Now putting the values of }x\text{ and }y.\\ {\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)\end{array}$

Thus the formula is proved.