Question

What signs are sec( – 140°) and cot( – 140°)? sec 140°) > 0 and cot( - 140°) < 0 sec- 140') < 0 and coté – 140°) > 0 They are both positive. They are both negative.

Answer 1

Given the trigonometry,

`\sec (-140^0)and\cot (-140^0)`

Solving for sec(-140°)

Express with cos

`\begin{gathered} \sec \mleft(-140^(\circ\: )\mright)=(1)/(\cos\left(-140^(\circ\:)\right)) \\ =(1)/(\cos\left(-140^(\circ\:)\right)) \end{gathered}`

Use the following property: cos(-x)=cos(x)

`\begin{gathered} \cos \mleft(-140^(\circ\: )\mright)=\cos \mleft(140^(\circ\: )\mright) \\ =(1)/(\cos\left(140^(\circ\:)\right))=(1)/(-0.7660)=-1.3054830\approx-1.31(2dp) \\ \therefore The\text{ answer is -1.31.} \end{gathered}`

Let us now solve for cot(-140°)

Express with tan

`\cot (-140^0)=(1)/(\tan (-140^0))`

Use the following property: tan(-x)=tan(x)

`\begin{gathered} \tan (-140^0)=\tan (140^0) \\ =(1)/(\tan(140^0))=(1)/(-0.8390)=-1.191895\approx-1.19(2\text{ decimal places)} \\ \text{The answer is -1}.19 \end{gathered}`

From above results gotten, we can conclude that both are negative.

Hence, the correct option is Option 4.