Answer
$\sin\theta$ = $- \frac{4}{\sqrt {17}}$
$\cos\theta$ =$ -\frac{1}{\sqrt {17}}$
$\tan\theta$ = 4
$\cot\theta$ =$ \frac{1}{4}$ (given)
$\sec\theta$ = $-\sqrt {17}$
$\csc\theta$ =$- \frac{\sqrt {17}}{4}$
Work Step by Step
Considering $\theta$ as reference angle, we may draw a right triangle ABC to calculate other trigonometric functions-
In right triangle ABC-
Given $\cot\theta$ =$ \frac{x}{y}$ = $ \frac{1}{4}$ (Ignoring sign)
Using Pythagorous theorem-
$r =\sqrt {x^{2}+y^{2}}$ =$ \sqrt {1^{2}+4^{2}}$ = $\sqrt {17}$
i.e. x= 1, y= 4 and r=$\sqrt {17}$
Given cot is positive and sin is negative, therefore $\theta$ lies in quadrant III, hence-
$\sin\theta$ =$ -\frac{y}{r}$ = $- \frac{4}{\sqrt {17}}$
$\cos\theta$ =$ -\frac{x}{r}$ =$ -\frac{1}{\sqrt {17}}$
$\tan\theta$ =$ \frac{y}{x}$ =$\frac{4}{1}$ = 4
$\cot\theta$ =$ \frac{x}{y}$ =$ \frac{1}{4}$ (given)
$\sec\theta$ =$ -\frac{r}{x}$ =$ -\frac{\sqrt {17}}{1}$ = $-\sqrt {17}$
$\csc\theta$ =$ -\frac{r}{y}$ =$- \frac{\sqrt {17}}{4}$
