Question

Si tan(a) =1, sec(b) = 2 y cotC = √3, calcule tan(b-a+c). ​

Answer 1

Respuesta:

Explicación paso a paso:

sec (b) = 1/cos(b) = 2

cos b = 1/2 = x/h = cateto/hipotenusa

y² = 2² - 1² = 4 - 1

y = $\sqrt{3}$

tan(b) = y/x = $\sqrt{3}$/1 = $\sqrt{3}$

cot (C) = 1/tan (C) = $\sqrt{3}$

tan (C) = 1/$\sqrt{3}$

Luego

$tan ( b+ c - a) = \frac{tan(b+c) + tan (a)}{1-tan(b+c)*tan(a)}$

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