Question

Calcular secx sabendo que senx = 2ab/ a^2 + b^2 com a>b>0

Answer 1

$sen(x)=\dfrac{2ab}{a^{2}+b^{2}}>0$
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Relação fundamental da trigonometria:

$sen^{2}(x)+cos^{2}(x)=1\\\\\\sen^{2}(x)+\dfrac{1}{sec^{2}(x)}=1\\\\\\\dfrac{1}{sec^{2}(x)}=1-sen^{2}(x)\\\\\\\dfrac{1}{sec^{2}(x)}=1-\left(\dfrac{2ab}{a^{2}+b^{2}}\right)^{2}\\\\\\\dfrac{1}{sec^{2}(x)}=1-\dfrac{4a^{2}b^{2}}{(a^{2}+b^{2})^{2}}\\\\\\\dfrac{1}{sec^{2}(x)}=\dfrac{(a^{2}+b^{2})^{2}-4a^{2}b^{2}}{(a^{2}+b^{2})^{2}}$

Expandindo (a² + b²)² no numerador:

$\dfrac{1}{sec^{2}(x)}=\dfrac{(a^{2})^{2}+2\cdot a^{2}\cdot b^{2}+(b^{2})^{2}-4a^{2}b^{2}}{(a^{2}+b^{2})^{2}}\\\\\\\dfrac{1}{sec^{2}(x)}=\dfrac{a^{4}+2a^{2}b^{2}-4a^{2}b^{2}+b^{4}}{(a^{2}+b^{2})^{2}}\\\\\\\dfrac{1}{sec^{2}(x)}=\dfrac{a^{4}-2a^{2}b^{2}+b^{4}}{(a^{2}+b^{2})^{2}}$

Temos o resultado de um quadrado da diferença no numerador:

$\dfrac{1}{sec^{2}(x)}=\dfrac{(a^{2}-b^{2})^{2}}{(a^{2}+b^{2})^{2}}\\\\\\sec^{2}(x)=\dfrac{(a^{2}+b^{2})^{2}}{(a^{2}-b^{2})^{2}}\\\\\\sec(x)=\pm\sqrt{\dfrac{(a^{2}+b^{2})^{2}}{(a^{2}-b^{2})^{2}}}\\\\\\\\\boxed{\boxed{sec(x)=\pm~\dfrac{a^{2}+b^{2}}{a^{2}-b^{2}}}}$