Question

Mostre que:

$\mathsf{\dfrac{1}{1+tg^2(\beta)}-\dfrac{1}{1+tg^2(\alpha)}=(sen(\alpha)-sen(\beta))\cdot(sen(\alpha)+sen(\beta))}$

___________

Responder de forma detalhada.

Answer 1

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}}$

Lembrando que $\mathsf{\text{tg}(\beta)=\dfrac{\text{sen}(\beta)}{\text{cos}(\beta)}}$ e $\mathsf{\text{tg}(\alpha)=\dfrac{\text{sen}(\alpha)}{\text{cos}(\alpha)}}$, obtemos:

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}=\dfrac{1}{1+\left[\dfrac{\text{sen}(\beta)}{\text{cos}(\beta)}\right]^2}-\dfrac{1}{1+\left[\dfrac{\text{sen}(\alpha)}{\text{cos}(\alpha)}\right]^2}}$

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}=\dfrac{1}{1+\dfrac{\text{sen}^2(\beta)}{\text{cos}^2(\beta)}}-\dfrac{1}{1+\dfrac{\text{sen}^2(\alpha)}{\text{cos}^2(\alpha)}}}$

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}=\dfrac{1}{\dfrac{\text{cos}^2(\beta)+\text{sen}^2(\beta)}{\text{cos}^2(\beta)}}-\dfrac{1}{\dfrac{\text{cos}^2(\alpha)+\text{sen}^2(\alpha)}{\text{cos}^2(\alpha)}}}$

Pela relação fundamental da trigonometria, temos que:

$\mathsf{\text{sen}^2(\beta)+\text{cos}^2(\beta)=\text{sen}^2(\alpha)+\text{cos}^2(\alpha)=1}$ 

Assim:

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}=\dfrac{1}{\dfrac{1}{\text{cos}^2(\beta)}}-\dfrac{1}{\dfrac{1}{\text{cos}^2(\alpha)}}}$

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}=\text{cos}^2(\beta)-\text{cos}^2(\alpha)}}$

Novamente pela relação fundamental, podemos afirmar que:

$\mathsf{\text{cos}^2(\beta)=1-\text{sen}^2(\beta)}$ 

$\mathsf{\text{cos}^2(\alpha)=1-\text{sen}^2(\alpha)}$. 

Logo:

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}=1-\text{sen}^2(\beta)-[1-\text{sen}^2(\alpha)]}$

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}=1-\text{sen}^2(\beta)-1+\text{sen}^2(\alpha)}$

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}=\text{sen}^2(\alpha)-\text{sen}^2(\beta)}$

Utilizando o produto notável $\mathsf{a^2-b^2=(a-b)\cdot(a+b)}$, obtemos:

$\mathsf{\dfrac{1}{1+\text{tg}^2(\beta)}-\dfrac{1}{1+\text{tg}^2(\alpha)}=[\text{sen}(\alpha)-\text{sen}(\beta)]\cdot[\text{sen}(\alpha)+\text{sen}(\beta)]}$