Given the trigonometry,
[tex]\sec (-140^0)and\cot (-140^0)[/tex]Solving for sec(-140°)
Express with cos
[tex]\begin{gathered} \sec \mleft(-140^{\circ\: }\mright)=\frac{1}{\cos\left(-140^{\circ\:}\right)} \\ =\frac{1}{\cos\left(-140^{\circ\:}\right)} \end{gathered}[/tex]Use the following property: cos(-x)=cos(x)
[tex]\begin{gathered} \cos \mleft(-140^{\circ\: }\mright)=\cos \mleft(140^{\circ\: }\mright) \\ =\frac{1}{\cos\left(140^{\circ\:}\right)}=\frac{1}{-0.7660}=-1.3054830\approx-1.31(2dp) \\ \therefore The\text{ answer is -1.31.} \end{gathered}[/tex]Let us now solve for cot(-140°)
Express with tan
[tex]\cot (-140^0)=\frac{1}{\tan (-140^0)}[/tex]Use the following property: tan(-x)=tan(x)
[tex]\begin{gathered} \tan (-140^0)=\tan (140^0) \\ =\frac{1}{\tan(140^0)}=\frac{1}{-0.8390}=-1.191895\approx-1.19(2\text{ decimal places)} \\ \text{The answer is -1}.19 \end{gathered}[/tex]From above results gotten, we can conclude that both are negative.
Hence, the correct option is Option 4.