Answer
$\displaystyle \cos t=\frac{3}{5}\quad \tan t=-\frac{4}{3}$
$\displaystyle \sec t=\frac{5}{3}\quad\csc t=-\frac{5}{4}\quad\cot t=-\frac{3}{4}$
Work Step by Step
$\sin^{2}t+\cos^{2}t=1\qquad\Rightarrow\cos t=\pm\sqrt{1-\sin^{2}t}$
In quadrant IV, cos and sec are positive,
all the other functions are negative (see table of signs, p.410)
So
$\displaystyle \cos t=+\sqrt{1-(-\frac{4}{5})^{2}}=\sqrt{\frac{25-16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$
$\displaystyle \tan t=\frac{\sin t}{\cos t}=\frac{-\frac{4}{5}}{\frac{3}{5}}=-\frac{4}{3}$
$\displaystyle \sec t=\frac{1}{\cos t}=\frac{5}{3}$
$\displaystyle \csc t=\frac{1}{\sin t}=-\frac{5}{4}$
$\displaystyle \cot t=\frac{1}{\tan t}=-\frac{3}{4}$
