From the figure (i) given below, calculate all the six t-ratios for both acute angles.

In right angle triangle ABC,

By pythagoras theorem we get :

⇒ AC² = AB² + BC²

⇒ 3² = AB² + 2²

⇒ 9 = AB² + 4

⇒ AB² = 9 - 4

⇒ AB² = 5

⇒ AB = $\sqrt{5}$.

For angle A,

$\Rightarrow \text{sin A} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{AC} = \dfrac{2}{3}. \\[1em] \Rightarrow \text{cos A} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{AB}{AC} = \dfrac{\sqrt{5}}{3}. \\[1em] \Rightarrow \text{tan A} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{BC}{AB} = \dfrac{2}{\sqrt{5}}. \\[1em] \Rightarrow \text{cot A} = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] = \dfrac{AB}{BC} = \dfrac{\sqrt{5}}{2}. \\[1em] \Rightarrow \text{sec A} = \dfrac{\text{Hypotenuse}}{\text{Base}} \\[1em] = \dfrac{AC}{AB} = \dfrac{3}{\sqrt{5}}. \\[1em] \Rightarrow \text{cosec A} = \dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \\[1em] = \dfrac{AC}{BC} = \dfrac{3}{2}. \\[1em]$

For angle C,

$\Rightarrow \text{sin C} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \dfrac{AB}{AC} = \dfrac{\sqrt{5}}{3}. \\[1em] \Rightarrow \text{cos C} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{AC} = \dfrac{2}{3}. \\[1em] \Rightarrow \text{tan C} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AB}{BC} = \dfrac{\sqrt{5}}{2}. \\[1em] \Rightarrow \text{cot C} = \dfrac{\text{Base}}{\text{Perpendicular}} \\[1em] = \dfrac{BC}{AB} = \dfrac{2}{\sqrt{5}}. \\[1em] \Rightarrow \text{sec C} = \dfrac{\text{Hypotenuse}}{\text{Base}} \\[1em] = \dfrac{AC}{BC} = \dfrac{3}{2}. \\[1em] \Rightarrow \text{cosec C} = \dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \\[1em] = \dfrac{AC}{AB} = \dfrac{3}{\sqrt{5}}. \\[1em]$