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

Answer:

Concept:

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

SOH CAH TOA

$\begin{gathered} SOH \\ sin\theta=\frac{opposite}{hypotenus}=S=\frac{O}{H} \\ \cos\theta=\frac{adjacent}{hypotenus}=C=\frac{A}{H} \\ \tan\theta=\frac{opposite}{adjacent}=T=\frac{O}{A} \end{gathered}$

Using the inverse trigonometric identity,

$\begin{gathered} cosecx^0=\frac{1}{sinx^0} \\ secx^0=\frac{1}{cosx^0} \\ cotx^0=\frac{1}{tanx^0} \end{gathered}$

By simplifying further, we will have that

$\begin{gathered} cosecx^0=\frac{1}{s\imaginaryI nx^{0}} \\ cosecx^0=\frac{1}{\frac{opposite}{hypotenu}}=1\times\frac{hypotenus}{opposite} \\ cosecx^0=\frac{hypotenus}{opposite} \end{gathered}$$\begin{gathered} secx^{0}=\frac{1}{cosx^{0}} \\ secx^0=\frac{1}{\frac{adjacent}{hypotenus}}=1\times\frac{hypotenus}{adjacent} \\ secx^0=\frac{hypotenus}{adjacent} \end{gathered}$$\begin{gathered} cotx^{0}=\frac{1}{tanx^{0}} \\ cotx^0=\frac{1}{\frac{opposite}{adjacent}}=1\times\frac{adjacent}{opposite} \\ cotx^0=\frac{adjacent}{oppos\imaginaryI te} \end{gathered}$

Hence,

The final answer is

$\Rightarrow secx^0=\frac{hypotenus}{adjacent}$