Answer
$\cos \theta = - \frac{4}{5}$
$\tan \theta = -\frac{3}{4}$
$\sec \theta = -\frac{5}{4}$
$\csc \theta = \frac{5}{3}$
$\cot \theta = -\frac{4}{3} $
Work Step by Step
We have the following information :
$\sin \theta = \frac{y}{r} = \frac{3}{5}$
$\cos \theta \lt 0$
From the first expression we can assume, that :
$y=3$
$r=5$
To find $y$, we can solve the following equation :
$r=\sqrt{x^2+y^2}$
$5=\sqrt{x^2+3^2}$
$25=x^2+9$
$x^2=16$
$x=±4$
From the given information, we know that $\cos \theta \lt 0$, so $x$ value has to be negative
$x = -4$
So, the remaining $5$ trigonometric functions will be :
$\cos \theta = \frac{x}{r} =- \frac{4}{5}$
$\tan \theta = \frac{y}{x} = -\frac{3}{4}$
$\sec \theta = \frac{r}{x} = -\frac{5}{4}$
$\csc \theta = \frac{r}{y} = \frac{5}{3}$
$\cot \theta = \frac{x}{y} = -\frac{4}{3} $
