Question

Tg (π-x) = ?

em que x ≠ 2 + kπ , k pertence aos inteiros

a resposta é -Tgx mas não sei como chegar​

Answer 1

Arco Subtração da tangente :

$\displaystyle \text{Tg}(\text a -\text b) = \frac{\text{Tg(a)}-\text{Tg(b)}}{1+\text{Tg(a).Tg(b) }}$

E sabemos que :

$\displaystyle \text{Tg}(\pi) = \frac{\text{Sen}(\pi)}{\text{Cos}(\pi)} = \frac{0}{1} = 0$

Temos :

$\text{Tg}(\pi - \text x)$

Abrindo o arco subtração

$\displaystyle \text{Tg}(\pi -\text x) = \frac{\text{Tg}(\pi)-\text{Tg(x)}}{1+\text{Tg}(\pi).\text{Tg(x)}}$

$\displaystyle \text{Tg}(\pi -\text x) = \frac{0-\text{Tg(x)}}{1+0.\text{Tg(x)}}$

$\displaystyle \text{Tg}(\pi -\text x) = \frac{-\text{Tg(x)}}{1}$

$\huge\boxed{\displaystyle \text{Tg}(\pi -\text x) = -\text{Tg(x)}}\checkmark$

Answer 2

$\sf tan \left(\pi -x\right)=\dfrac{tan \left(\pi \right)-tan \left(x\right)}{1+tan \left(\ \pi \right)tan \left(x\right)}\\\\\\{Manipule\:o\:lado\:direito}\\\\\\tan \left(\pi -x\right)\\\\\\=-\dfrac{sin \left(x\right)}{cos \left(x\right)}\\\\\\{Manipule\:o\:lado\:esquerdo}\\\\\\\dfrac{tan \left(\pi \right)-tan \left(x\right)}{1+tan \left(\pi \right)tan \left(x\right)}\\\\\\Expresse \ com \ Seno , cosseno\\\\\\=-\dfrac{sin \left(x\right)}{cos \left(x\right)}$

$\sf {Então \: Podemos \: Afirmar \:que\:os\:dois\:lados\:podem\:adquirir\:a\:mesma\:forma}\\\\\\ \boxed{\sf \Rightarrow {Verdadeiro}}$