Question

Sabendo-se que x∈IV Quadrante e que sec⁡x=√2, determine o valor da expressão: A=(1+tan⁡x+csc⁡x)/(1+cot⁡x-csc⁡x )

Answer 1

$\mathrm{\sec{x}=\sqrt{2}\ \to\ 1+\tan^2{x}=\sec^2{x}\ \to\ 1+\tan^2{x}=(\sqrt{2})^2}\\ \mathrm{\tan^2{x}=2-1\ \to\ \tan^2{x}=1\ \to\ \tan{x}=\pm1\ \to\ \boxed{\mathrm{\tan{x}=-1}}}\\\\\\ \mathrm{\cot{x}=\dfrac{1}{\tan{x}}\ \to\ \cot{x}=\dfrac{1}{-1}\ \to\ \boxed{\mathrm{\cot{x}=-1}}}\\\\\\ \mathrm{1+\cot^2{x}=\csc^2{x}\ \to\ 1+(-1)^2=\csc^2{x}\ \to\ \csc^2{x}=1+1}\\ \mathrm{\csc^2{x}=2\ \to\ \csc{x}=\pm\sqrt{2}\ \to\ \boxed{\mathrm{\csc{x}=-\sqrt{2}}}}$

$\mathrm{A=\dfrac{1+\tan{x}+\csc{x}}{1+\cot{x}-\csc{x}}\ \to\ A=\dfrac{1+(-1)+(-\sqrt{2})}{1+(-1)-(-\sqrt{2})}}\\\\ \mathrm{A=\dfrac{1-1-\sqrt{2}}{1-1+\sqrt{2}}\ \to\ A=-\dfrac{\sqrt{2}}{\sqrt{2}}\ \to\ \boxed{\boxed{\mathbf{A=-1}}}}$