Question

Which equation is correct? (6 points)Group of answer choicessec x° = opposite ÷ adjacentcot x° = opposite ÷ adjacentcosec x° = opposite ÷ adjacentsec x° = hypotenuse ÷ adjacent

Answer 1

Concept:

To figure this question out, we will use the trigonometric ratios below

SOH CAH TOA

`\begin{gathered} SOH \\ sin\theta=(opposite)/(hypotenus)=S=(O)/(H) \\ \cos\theta=(adjacent)/(hypotenus)=C=(A)/(H) \\ \tan\theta=(opposite)/(adjacent)=T=(O)/(A) \end{gathered}`

Using the inverse trigonometric identity,

`\begin{gathered} cosecx^0=(1)/(sinx^0) \\ secx^0=(1)/(cosx^0) \\ cotx^0=(1)/(tanx^0) \end{gathered}`

By simplifying further, we will have that

`\begin{gathered} cosecx^0=(1)/(s\imaginaryI nx^(0)) \\ cosecx^0=(1)/((opposite)/(hypotenu))=1*(hypotenus)/(opposite) \\ cosecx^0=(hypotenus)/(opposite) \end{gathered}`
`\begin{gathered} secx^(0)=(1)/(cosx^(0)) \\ secx^0=(1)/((adjacent)/(hypotenus))=1*(hypotenus)/(adjacent) \\ secx^0=(hypotenus)/(adjacent) \end{gathered}`
`\begin{gathered} cotx^(0)=(1)/(tanx^(0)) \\ cotx^0=(1)/((opposite)/(adjacent))=1*(adjacent)/(opposite) \\ cotx^0=(adjacent)/(oppos\imaginaryI te) \end{gathered}`

Hence,

The final answer is

`\Rightarrow secx^0=(hypotenus)/(adjacent)`