Example 001 | Sum and Difference of Two Angles

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

$\tan (x + y) = \dfrac{\frac{1}{3} + \frac{1}{2}}{1 - \frac{1}{3}(\frac{1}{2})}$

$\tan (x + y) = \dfrac{\,\frac{5}{6}\,}{\frac{5}{6}}$

$\tan (x + y) = 1$           answer
 

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

$\tan (x - y) = \dfrac{\frac{1}{3} - \frac{1}{2}}{1 + \frac{1}{3}(\frac{1}{2})}$

$\tan (x - y) = \dfrac{\,-\frac{1}{6}\,}{\frac{7}{6}}$

$\tan (x - y) = -\frac{1}{7}$           answer

$\tan x = \frac{1}{3}$

$x = \arctan (\frac{1}{3})$

$x = 18.43^\circ$
 

$\tan y = \frac{1}{2}$

$y = \arctan (\frac{1}{2})$

$y = 26.57^\circ$
 

$\tan (x + y) = \tan (18.43^\circ + 26.57^\circ)$

$\tan (x + y) = \tan 45^\circ$

$\tan (x + y) = 1$           answer
 

$\tan (x - y) = \tan (18.43^\circ - 26.57^\circ)$

$\tan (x - y) = \tan (-8.14^\circ)$

$\tan (x - y) = -0.143$           answer