Use the given values to evaluate (if possible) three trigonometric functions cos x, csc x, tan x.
$$\sin x = \frac { 1 } { 5 }$$ , $$\cos x 0$$
A) $\begin{array} { l }
\cos x = - \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = 5 \
\tan x = \frac { \sqrt { 6 } } { 12 }
\end{array}$
B) $\begin{array} { l }
\cos x = \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = 5 \
\tan x = - \frac { \sqrt { 6 } } { 12 }
\end{array}$
C) $\begin{array} { l }
\cos x = - \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = - 5 \
\tan x = - \frac { \sqrt { 6 } } { 12 }
\end{array}$
D) $\begin{array} { l }
\cos x = \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = 5 \
\tan x = \frac { \sqrt { 6 } } { 12 }
\end{array}$
E) $\begin{array} { l }
\cos x = \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = \frac { 1 } { 5 } \
\tan x = \frac { \sqrt { 6 } } { 12 }
\end{array}$

Question

Use the given values to evaluate (if possible) three trigonometric functions cos x, csc x, tan x.
 $$\sin x = \frac { 1 } { 5 }$$ , $$\cos x > 0$$  
A)  $\begin{array} { l } 
\cos x = - \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = 5 \
\tan x = \frac { \sqrt { 6 } } { 12 }
\end{array}$  
B)  $\begin{array} { l } 
\cos x = \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = 5 \
\tan x = - \frac { \sqrt { 6 } } { 12 }
\end{array}$  
C)  $\begin{array} { l } 
\cos x = - \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = - 5 \
\tan x = - \frac { \sqrt { 6 } } { 12 }
\end{array}$  
D)  $\begin{array} { l } 
\cos x = \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = 5 \
\tan x = \frac { \sqrt { 6 } } { 12 }
\end{array}$  
E)  $\begin{array} { l } 
\cos x = \frac { 2 \sqrt { 6 } } { 5 } \
\csc x = \frac { 1 } { 5 } \
\tan x = \frac { \sqrt { 6 } } { 12 }
\end{array}$

Quizplus · Accepted Answer

Given $\sin x = \frac{1}{5}$, we can find $\cos x$ using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$. Solving for $\cos x$, we get $\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \left(\frac{1}{5}\right)^2} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$. Since $\cos x > 0$, we choose the positive root. The $\csc x$ is the reciprocal of $\sin x$, so $\csc x = \frac{1}{\sin x} = 5$. For $\tan x = \frac{\sin x}{\cos x} = \frac{1/5}{2\sqrt{6}/5} = \frac{\sqrt{6}}{12}$.

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