Answer
$\sec\theta=-4$
$\cos\theta=-\frac{1}{4}$
$ \sin\theta=\frac{\sqrt{ 15}}{4}$
$\csc\theta=\frac{4\sqrt {15}}{15}$
$\tan\theta=-\sqrt {15}$
$\cot\theta=-\frac{\sqrt {15}}{15}$
Work Step by Step
$\sec\theta=-4$
$\cos\theta=\frac{1}{\sec\theta}=\frac{1}{-4}=-\frac{1}{4}$
$\sin^{2}\theta=1-\cos^{2}\theta=1-(-\frac{1}{4})^{2}=\frac{15}{16}$
$\implies \sin\theta=\pm\sqrt {\frac{15}{16}}=\frac{\sqrt{ 15}}{4}$ (given $\sin\theta\gt0$)
$\csc\theta=\frac{1}{\sin\theta}=\frac{4}{\sqrt {15}}=\frac{4\sqrt {15}}{15}$
$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{\sqrt {15}}{4}}{-\frac{1}{4}}=-\sqrt {15}$
$\cot\theta=\frac{1}{\tan\theta}=-\frac{1}{\sqrt {15}}=-\frac{\sqrt {15}}{15}$
