## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\sec{\theta} =-\dfrac{5}{4}$ $\cos{\theta} =-\dfrac{4}{5}$ $\sin{\theta}= -\dfrac{3}{5}$ $\csc{\theta} =-\dfrac{5 }{3}$ $\cot{\theta} =\dfrac{4}{3}$
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, $\sin{\theta}= -\dfrac{4}{5} \times \dfrac{3}{4} = -\dfrac{3}{5}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ $\csc{\theta} = \dfrac{1}{-\dfrac{3}{5}}= -\dfrac{5 }{3}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{\dfrac{3}{4}} = \dfrac{4}{3}$