From the figure (2) given below, find the value of :

(i) sin x

(ii) cot x

(iii) cot² x - cosec² x

(iv) sec y

(v) tan² y - $\dfrac{1}{\text{cos}^2 y}$

In right angled ∆ABD,

By pythagoras theorem we get,

⇒ AD² = AB² + BD²

⇒ AD² = 3² + 4²

⇒ AD² = 9 + 16

⇒ AD² = 25

⇒ AD = $\sqrt{25}$

⇒ AD = 5

In right angled triangle DBC,

By pythagoras theorem we get,

⇒ DC² = DB² + BC²

⇒ 12² = 4² + BC²

⇒ BC² = 12² - 4²

⇒ BC² = 144 - 16

⇒ BC² = 128

⇒ BC = $\sqrt{128}$

⇒ BC = $8\sqrt{2}$.

(i) sin x = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{BD}{AD} = \dfrac{4}{5}$.

Hence, sin x = $\dfrac{4}{5}$.

(ii) cot x = $\dfrac{\text{Base}}{\text{Perpendicular}}$

= $\dfrac{AB}{BD} = \dfrac{3}{4}$.

Hence, cot x = $\dfrac{3}{4}$.

(iii) cosec x = $\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}$

= $\dfrac{AD}{BD} = \dfrac{5}{4}$.

Substituting values in cot² x - cosec² x we get :

$= \Big(\dfrac{3}{4}\Big)^2 - \Big(\dfrac{5}{4}\Big)^2 \\[1em] = \dfrac{9}{16} - \dfrac{25}{16} \\[1em] = -\dfrac{16}{16} \\[1em] = -1.$

Hence, cot² x - cosec² x = -1.

(iv) sec y = $\dfrac{\text{Hypotenuse}}{\text{Base}}$

= $\dfrac{CD}{BC} = \dfrac{12}{8\sqrt{2}} = \dfrac{3}{2\sqrt{2}}$.

Hence, sec y = $\dfrac{3}{2\sqrt{2}}$.

(v) tan y = $\dfrac{\text{Perpendicular}}{\text{Base}}$

= $\dfrac{BD}{BC} = \dfrac{4}{8\sqrt{2}} = \dfrac{1}{2\sqrt{2}}$.

Given,

$\Rightarrow \text{tan}^2 y - \dfrac{1}{\text{cos}^2 y} \\[1em] \Rightarrow \Big(\dfrac{1}{2\sqrt{2}}\Big)^2 - \text{sec}^2 y \\[1em] \Rightarrow \dfrac{1}{8} - \Big(\dfrac{3}{2\sqrt{2}}\Big)^2 \\[1em] \Rightarrow \dfrac{1}{8} - \dfrac{9}{8} \\[1em] \Rightarrow \dfrac{1 - 9}{8} \\[1em] \Rightarrow \dfrac{-8}{8} \\[1em] \Rightarrow -1.$

Hence, $\text{tan}^2 y - \dfrac{1}{\text{cos}^2 y} = -1$