Answer
$\tan{\theta}=\dfrac{3}{4}$
$\sec{\theta} =-\dfrac{5}{4}$
$\cos{\theta} =-\dfrac{4}{5}$
$\sin{\theta}=-\dfrac{3}{5}$
$\csc{\theta} = -\dfrac{5 }{3}$
Work Step by Step
Since $\tan{\theta}= \dfrac{1}{\cot{\theta}}$, then
$\tan{\theta}= \dfrac{1}{\frac{4}{3}} = \dfrac{3}{4}$
With $\sec{\theta} = \dfrac{1}{\cos{\theta}} \hspace{5pt} \text{ and } \hspace{5pt} \cos{\theta} <0$, then $\sec{\theta} <0$.
Note that $1+\tan^2{\theta} = \sec^2{\theta}$.
$\therefore \sec{\theta} = - \sqrt{1+\tan^2{\theta}}$
Thus,
$\sec{\theta} = - \sqrt{1+\left(\dfrac{3}{4} \right)^2} =-\dfrac{5}{4}$
$\cos{\theta} = \dfrac{1}{\sec{\theta}}$
$\cos{\theta} =\dfrac{1}{-\dfrac{5}{4}} = -\dfrac{4}{5}$
Since $\sin{\theta} = \cos{\theta} \tan{\theta}$, then
$\sin{\theta}= -\dfrac{4}{5} \times \dfrac{3}{4} = -\dfrac{3}{5}$
$\csc{\theta} = \dfrac{1}{\sin{\theta}}$
$\csc{\theta} = \dfrac{1}{-\dfrac{3}{5}}= -\dfrac{5 }{3}$
