Answer
$\sin\theta$ = $\frac{3\sqrt {5}}{7}$
$\cos\theta$ =$ -\frac{2}{7}$ (given)
$\tan\theta$ =$- \frac{3\sqrt {5}}{2}$
$\cot\theta$ =$- \frac{2}{3\sqrt {5}}$
$\sec\theta$ =$ -\frac{7}{2}$
$\csc\theta$ =$\frac{7}{3\sqrt {5}}$
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 $\cos\theta$ =$ \frac{x}{r}$ = $ \frac{2}{7}$ (Ignoring sign)
Using Pythagorous theorem-
$y =\sqrt {r^{2}-x^{2}}$ =$ \sqrt {7^{2}-2^{2}}$ = $\sqrt {45}$ = $3\sqrt {5}$
i.e. x= 2, y= $3\sqrt {5}$ and r=7
Given cos is negative and tan is negative, therefore $\theta$ lies in quadrant II, hence-
$\sin\theta$ =$ \frac{y}{r}$ = $\frac{3\sqrt {5}}{7}$
$\cos\theta$ =$ -\frac{x}{r}$ =$ -\frac{2}{7}$ (given)
$\tan\theta$ =$ -\frac{y}{x}$ =$- \frac{3\sqrt {5}}{2}$
$\cot\theta$ =$ -\frac{x}{y}$ =$- \frac{2}{3\sqrt {5}}$
$\sec\theta$ =$ -\frac{r}{x}$ =$ -\frac{7}{2}$
$\csc\theta$ =$ \frac{r}{y}$ =$\frac{7}{3\sqrt {5}}$
