Question

(UFSC)Se sen x=35e x ∈ 0 < x < $\pi$/2, calcule o valor numérico da expressão:

Answer 1

Resposta:

$(\frac{sec^{2}x.cotgx-cossecx.tgx}{6.senx.cossec^{2}x})^{-1}=\\\\(\frac{\frac{1}{cos^{2}x}.\frac{1}{tgx} -\frac{1}{senx}.\frac{senx}{cosx} }{6.senx.\frac{1}{sen^{2}x}} )^{-1}=\\\\(\frac{\frac{1}{cos^{2}x}.\frac{cosx}{senx}-\frac{1}{cosx}}{\frac{6}{senx}} )^{-1}=\\\\(\frac{\frac{1}{senxcosx}-\frac{1}{cosx}}{\frac{6}{senx}} )^{-1}=((\frac{1}{senxcosx}-\frac{1}{cosx}).{\frac{senx}{6}})^{-1}=((\frac{1}{cosx}-\frac{senx}{cosx}).{\frac{1}{6}})^{-1}=((\frac{1-senx}{cosx}).{\frac{1}{6}})^{-1}$

$sen^{2}x+cos^{2}x=1 => cos^{2}x=1-sen^{2}x=1-(\frac{3}{5})^{2}=\frac{25-9}{25} =\frac{16}{25} \\\\cosx=\pm\frac{4}{5}$

Como $0<x<\frac{\pi}{2}$ então x está no I Quadrante nesse caso cosx>0, portanto:$cosx=\frac{4}{5}$

$(\frac{sec^{2}x.cotgx-cossecx.tgx}{6.senx.cossec^{2}x})^{-1}=((\frac{1-senx}{cosx}).{\frac{1}{6}})^{-1}=((\frac{1-\frac{3}{5}}{\frac{4}{5}}).{\frac{1}{6}})^{-1}=((\frac{\frac{2}{5}}{\frac{4}{5}}).{\frac{1}{6}})^{-1}=((\frac{2}{5}.{\frac{5}{4}}).{\frac{1}{6}})^{-1}=(\frac{2}{4}.{\frac{1}{6}})^{-1}=(\frac{1}{12})^{-1}=12$