Question

Find the trigonometric ratios in standard position whose terminal arm passes through the points.
(vi) $\left(-2,-3\right)$

Answer 1

Let the terminal arm pass through point F$\left(-2,-3\right)$.
$\therefore x=-2\mathrm{and}y=-3$

$$\begin{gathered} \mathrm{Here},r=\sqrt{x^2+y^2}=\sqrt{{\left(-2\right)}^2+{\left(-3\right)}^2}=\sqrt{13} \\ \mathrm{Now},\mathrm{let}\mathrm{the}\mathrm{angle}\mathrm{be}\theta . \\ \therefore \sin \theta =\frac{y}{r}=-\frac{3}{\sqrt{13}};\cos ec\theta =\frac{r}{y}=-\frac{\sqrt{13}}{3} \\ \cos \theta =\frac{x}{r}=-\frac{2}{\sqrt{13}};sec\theta =\frac{r}{x}=-\frac{\sqrt{13}}{2} \\ \tan \theta =\frac{y}{x}=\frac{3}{2};cot\theta =\frac{x}{y}=\frac{2}{3} \end{gathered}$$