Question

Sabendo que tga = 1/5 e tgb = 1/10 calcule tg (a-b) (com cálculo, por favor)

Answer 1

Explicação passo-a-passo:

$\sf tg~(a-b)=\dfrac{tg~a-tg~b}{1+tg~a\cdot tg~b}$

$\sf tg~(a-b)=\dfrac{\frac{1}{5}-\frac{1}{10}}{1+\frac{1}{5}\cdot\frac{1}{10}}$

$\sf tg~(a-b)=\dfrac{\frac{2-1}{10}}{1+\frac{1}{50}}$

$\sf tg~(a-b)=\dfrac{\frac{1}{10}}{1+\frac{1}{50}}$

$\sf tg~(a-b)=\dfrac{\frac{1}{10}}{\frac{50+1}{50}}$

$\sf tg~(a-b)=\dfrac{\frac{1}{10}}{\frac{51}{50}}$

$\sf tg~(a-b)=\dfrac{1}{10}\cdot\dfrac{50}{51}$

$\sf tg~(a-b)=\dfrac{50}{510}$

$\sf tg~(a-b)=\dfrac{5}{51}$

Answer 2

Resposta:

g (a−b)=

1+tg a⋅tg b

tg a−tg b

\sf tg~(a-b)=\dfrac{\frac{1}{5}-\frac{1}{10}}{1+\frac{1}{5}\cdot\frac{1}{10}}tg (a−b)=

1+

5

1

⋅

10

1

5

1

−

10

1

\sf tg~(a-b)=\dfrac{\frac{2-1}{10}}{1+\frac{1}{50}}tg (a−b)=

1+

50

1

10

2−1

\sf tg~(a-b)=\dfrac{\frac{1}{10}}{1+\frac{1}{50}}tg (a−b)=

1+

50

1

10

1

\sf tg~(a-b)=\dfrac{\frac{1}{10}}{\frac{50+1}{50}}tg (a−b)=

50

50+1

10

1

\sf tg~(a-b)=\dfrac{\frac{1}{10}}{\frac{51}{50}}tg (a−b)=

50

51

10

1

\sf tg~(a-b)=\dfrac{1}{10}\cdot\dfrac{50}{51}tg (a−b)=

10

1

⋅

51

50

\sf tg~(a-b)=\dfrac{50}{510}tg (a−b)=

510

50

\sf tg~(a-b)=\dfrac{5}{51}tg (a−b)=

51

5

Explicação passo-a-passo:

denada