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

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