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