Question

Find the trigonometric ratios in standard position whose terminal arm passes through the points.
(iv) $\left(-1,\sqrt{3}\right)$

Answer 1

Let the terminal arm pass through the point D$\left(-1,\sqrt{3}\right)$.
$$\begin{gathered} \therefore x=-1\mathrm{and}y=\sqrt{3} \\ \mathrm{Here},r=\sqrt{x^2+y^2}=\sqrt{{\left(-1\right)}^2+(\sqrt{3}{)}^2}=2 \\ \mathrm{Now},\mathrm{let}\mathrm{the}\mathrm{angle}\mathrm{be}\theta . \\ \therefore \sin \theta =\frac{y}{r}=\frac{\sqrt{3}}{2};\cos ec\theta =\frac{r}{y}=\frac{2}{\sqrt{3}} \\ \cos \theta =\frac{x}{r}=\frac{-1}{2}=-\frac{1}{2};sec\theta =\frac{r}{x}=\frac{2}{-1}=-2 \\ \tan \theta =\frac{y}{x}=\frac{\sqrt{3}}{-1}=-\sqrt{3};cot\theta =\frac{x}{y}=\frac{-1}{\sqrt{3}}=-\frac{1}{\sqrt{3}} \end{gathered}$$