Question

$\begin{array}{l}\mathsf{Calcular~sec(x)~sabendo~que~sen(x)=\dfrac{2ab}{a^2+b^2}~com~a\ \textgreater \ b\ \textgreater \ 0.}\\\\\\\mathsf{gab:sec(x)=\pm~\dfrac{a^2+b^2}{a^2-b^2} }\end{array}$

Answer 1

Pela relação fundamental da trigonometria,

$\mathtt{sen}^{2}\theta+\mathtt{cos}^{2}\theta=1~~~~\forall~\theta\in\mathbb{R}$

Portanto

$\mathtt{cos}^{2}x=1-\mathtt{sen}^{2}x\\\\\\\mathtt{cos}^{2}x=1-\bigg[\dfrac{2ab}{a^{2}+b^{2}}\bigg]^{2}\\\\\\\mathtt{cos}^{2}x=1-\dfrac{(2ab)^{2}}{(a^{2}+b^{2})^{2}}\\\\\\\mathtt{cos}^{2}x=\dfrac{(a^{2}+b^{2})^{2}}{(a^{2}+b^{2})^{2}}-\dfrac{(2ab)^{2}}{(a^{2}+b^{2})^{2}}\\\\\\\mathtt{cos}^{2}x=\dfrac{(a^{2}+b^{2})^{2}-(2ab)^{2}}{(a^{2}+b^{2})^{2}}$

Usando $p^{2}-q^{2}=(p+q)(p-q)$:

$\mathtt{cos}^{2}x=\dfrac{(a^{2}+b^{2}+2ab)(a^{2}+b^{2}-2ab)}{(a^{2}+b^{2})^{2}}\\\\\\\mathtt{cos}^{2}x=\dfrac{(a^{2}+2ab+b^{2})(a^{2}-2ab+b^{2})}{(a^{2}+b^{2})^{2}}\\\\\\\mathtt{cos}^{2}x=\dfrac{(a+b)^{2}(a-b)^{2}}{(a^{2}+b^{2})^{2}}$

Como $\sec^{2}x=\frac{1}{\cos^{2}x}$:

$\sec^{2}x=\dfrac{1}{\cos^{2}x}=\dfrac{(a^{2}+b^{2})^{2}}{(a+b)^{2}(a-b)^{2}}$

Tirando a raiz de ambos os lados:

$\sqrt{\sec^{2}x}=\sqrt{\dfrac{(a^{2}+b^{2})^{2}}{(a+b)^{2}(a-b)^{2}}}\\\\\\|\sec\,x|=\dfrac{\sqrt{(a^{2}+b^{2})^{2}}}{\sqrt{(a+b)^{2}(a-b)^{2}}}$

Mas,
 
$\bullet \,\,(a,b)\in\mathbb{R}^{2}~\Rightarrow~a^{2}+b^{2}\ge0~\Rightarrow~\sqrt{(a^{2}+b^{2})^{2}}=|a^{2}+b^{2}|=a^{2}+b^{2}\\\\\bullet\,\,a~\textgreater~b~\textgreater~0~\Rightarrow~a+b~\textgreater~0~\Rightarrow~\sqrt{(a+b)^{2}}=|a+b|=a+b\\\\\bullet\,\,a~\textgreater~b~\Rightarrow~a-b~\textgreater~0~\Rightarrow~\sqrt{(a-b)^{2}}=|a-b|=a-b$

Então

$|sec\,x|=\dfrac{a^{2}+b^{2}}{(a+b)(a-b)}\\\\\\|\sec\,x|=\dfrac{a^{2}+b^{2}}{a^{2}-b^{2}}\\\\\\\boxed{\boxed{\sec\,x=\pm\,\,\dfrac{a^{2}+b^{2}}{a^{2}-b^{2}}}}$