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

Question

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

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 $\cos x$ gives $\cos x = \pm \sqrt{1 - \sin^2 x} = \pm \sqrt{1 - \left(\frac{1}{5}\right)^2} = \pm \sqrt{\frac{24}{25}} = \pm \frac{2\sqrt{6}}{5}$. Since the sign of $\cos x$ is not specified, we assume the positive value for the principal value. The $\csc x$ is the reciprocal of $\sin x$, so $\csc x = \frac{1}{\sin x} = 5$. The $\tan x$ is the ratio of $\sin x$ to $\cos x$, so $\tan x = \frac{\sin x}{\cos x} = \frac{1/5}{2\sqrt{6}/5} = \frac{1}{2\sqrt{6}} = \frac{\sqrt{6}}{12}$.

Use the Given Values to Evaluate (If Possible) Three Trigonometric sin⁡x=15\sin x = \frac { 1 } { 5 }sinx=51​

Use the Given Values to Evaluate (If Possible) Three Trigonometric $\sin x = \frac { 1 } { 5 }$