Answer
$\sin{\theta} =- \dfrac{\sqrt{15}}{4}$
$\tan{\theta}=\sqrt{15}$
$\csc{\theta} =-\dfrac{4 \sqrt{15} }{15}$
$\sec{\theta} =-4$
$\cot{\theta} =\dfrac{\sqrt{15}}{15}$
Work Step by Step
$\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} >0$
Since $\cos{\theta} \lt 0$, then $\sin{\theta}$ must be negative.
Thus,
$\sin{\theta} = -\sqrt{1-\cos^2 {\theta}}$
$\sin{\theta} = -\sqrt{1-\left(-\dfrac{1}{4} \right)^2}= - \dfrac{\sqrt{15}}{4}$
$\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$
$\tan{\theta}= \dfrac{- \dfrac{\sqrt{15}}{4}}{-\dfrac{1}{4}} = \sqrt{15}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}}$
$\csc{\theta} = \dfrac{1}{- \dfrac{\sqrt{15}}{4}}= -\dfrac{4 \sqrt{15} }{15}$
$\sec{\theta} = \dfrac{1}{\cos{\theta}}$
$\sec{\theta} = \dfrac{1}{-\dfrac{1}{4}} = -4$
$\cot{\theta} = \dfrac{1}{\tan{\theta}}$
$\cot{\theta} = \dfrac{1}{\sqrt{15}} = \dfrac{\sqrt{15}}{15}$
