Answer:
[tex]\tan (30 + 60) = unde f ined[/tex]
Step-by-step explanation:
Given
[tex]\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A * \tan B}[/tex] ---- right expression
[tex]A = 30^o; B = 60^o[/tex]
Required
Solve
[tex]\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A * \tan B}[/tex]
Substitute: [tex]A = 30^o; B = 60^o[/tex]
[tex]\tan (30 + 60) = \frac{\tan 30 + \tan 60}{1 - \tan 30 * \tan 60}[/tex]
In radical forms, we have:
[tex]\tan (30 + 60) = \frac{\sqrt 3/3 + \sqrt 3}{1 - \sqrt 3/3 * \sqrt 3}[/tex]
[tex]\tan (30 + 60) = \frac{\sqrt 3/3 + \sqrt 3}{1 - 1}[/tex]
[tex]\tan (30 + 60) = \frac{\sqrt 3/3 + \sqrt 3}{0}[/tex]
[tex]\tan (30 + 60) = unde f ined[/tex]
