Question

Find the trigonometric ratios in standard position whose terminal arm passes through the points.
(v) $\left(1,-1\right)$

Answer 1

Let the terminal arm pass through point E$\left(1,-1\right)$.
$\therefore x=1\mathrm{and}y=-1$

$$\begin{gathered} \mathrm{Here},r=\sqrt{x^2+y^2}=\sqrt{1^2+(-1{)}^2}=\sqrt{2} \\ \mathrm{Now},\mathrm{let}\mathrm{the}\mathrm{angle}\mathrm{be}\theta . \\ \therefore \sin \mathrm{\theta }=\frac{\mathrm{y}}{\mathrm{r}}=\frac{-1}{\sqrt{2}};\mathrm{cosec}\mathrm{\theta }=\frac{\mathrm{r}}{\mathrm{y}}=\frac{\sqrt{2}}{-1}=-\sqrt{2} \\ \cos \mathrm{\theta }=\frac{\mathrm{x}}{\mathrm{r}}=\frac{1}{\sqrt{2}};\sec \mathrm{\theta }=\frac{\mathrm{r}}{\mathrm{x}}=\frac{\sqrt{2}}{1}=\sqrt{2} \\ \tan \mathrm{\theta }=\frac{\mathrm{y}}{\mathrm{x}}=\frac{-1}{1}=-1;\cot \mathrm{\theta }=\frac{\mathrm{x}}{\mathrm{y}}=\frac{1}{-1}=-1 \end{gathered}$$