# Thread: counting by two methods

1. ## counting by two methods

reals numbers.
1 ) Calculate :

2 ) Let :

Calculate by anathor mhetode the real number E .

2. Gello, dhiab!

I hope I understand the problem . . .

We should know this identity: .$\displaystyle \tan(\alpha - \beta) \;=\;\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}$

Given: .$\displaystyle \begin{array}{c}a \:=\:\tan A \\ b \:=\:\tan B \\ x \:=\:\tan X\end{array}$

Calculate: . $\displaystyle E \;=\;\frac{\dfrac{x-a}{1+ax} - \dfrac{x-b}{1+bx}} {1 + \dfrac{x-a}{1+ax}\cdot\dfrac{x-b}{1+bx}}$

We have: . $\displaystyle E \;=\;\frac{\dfrac{\tan X - \tan A}{1 + \tan A\tan X} - \dfrac{\tan X - \tan B}{1 + \tan B\tan X} }{1 + \dfrac{\tan X - \tan A}{1 + \tan A\tan X}\cdot\dfrac{\tan X - \tan B}{1 + \tan B\tan X}}$

. . . . . . . . $\displaystyle E \;=\;\frac{\tan(X-A) - \tan(X-B)}{1 + \tan(X-A)\tan(X-B)}$

. . . . . . . . $\displaystyle E \;=\;\tan\bigg[(X-A) - (X-B)\bigg]$

. . . . . . . . $\displaystyle E \;=\;\tan(B-A)$