Respuesta :

Banabanana

Answer:

[tex]\tt B. \ \ \ \cfrac{1+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}[/tex]

Step-by-step explanation:

[tex]\displaystyle\tt \tan75^o=\tan(45^o+30^o)=\frac{\tan45^o+\tan30^o}{1-\tan45^o\cdot\tan30^o} =\frac{1+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}[/tex]

carlosego

For this case we have to define that:

[tex]tg (x + y) = \frac {tg (x) + tg (y)} {1-tg (x) * tg (y)}[/tex]

So, according to the problem we have:

[tex]tg (45 + 30) = \frac {tg (45) + tg (30)} {1-tg (45) * tg (30)}[/tex]

By definition we have to:

[tex]tg (45) = 1\\tg (30) = \frac {\sqrt {3}} {3}[/tex]

Substituting we have:

[tex]tg (45 + 30) = \frac {1+ \frac {\sqrt {3}} {3}} {1-1 * \frac {\sqrt {3}} {3}}\\tg (45 + 30) = \frac {1+ \frac {\sqrt {3}} {3}} {1- \frac {\sqrt {3}} {3}}[/tex]

Answer:

option B

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