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

(i) tan x

(ii) cos y

(iii) cosec² y - cot² y

(iv) $\dfrac{5}{\text{sin x}} + \dfrac{3}{\text{sin y}} - 3\text{ cot y}$.

From Figure,

BD = BC - CD = 21 - 5 = 16.

In right-angled ∆ACD,

By pythagoras theorem we get,

⇒ AC² = AD² + CD²

⇒ AD² = AC² - CD²

⇒ AD² = (13)² - (5)²

⇒ AD² = 169 - 25

⇒ AD² = 144

⇒ AD = $\sqrt{144}$

⇒ AD = 12.

In right-angled ∆ABD,

By pythagoras theorem we get,

⇒ AB² = AD² + BD²

⇒ AB² = 12² + 16²

⇒ AB² = 144 + 256

⇒ AB² = 400

⇒ AB = $\sqrt{400}$

⇒ AB = 20.

(i) In right-angled ∆ACD,

tan x = $\dfrac{\text{Perpendicular}}{\text{Base}}$

= $\dfrac{CD}{AD} = \dfrac{5}{12}$.

Hence, tan x = $\dfrac{5}{12}$.

(ii) In right-angled ∆ABD,

cos y = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

= $\dfrac{BD}{AB} = \dfrac{16}{20} = \dfrac{4}{5}$.

Hence, cos y = $\dfrac{4}{5}$.

(iii) In right-angled ∆ABD,

cosec y = $\dfrac{\text{Hypotenuse}}{\text{Perpendicular}}$

= $\dfrac{AB}{AD} = \dfrac{20}{12} = \dfrac{5}{3}$.

cot y = $\dfrac{\text{Base}}{\text{Perpendicular}}$

= $\dfrac{BD}{AD} = \dfrac{16}{12} = \dfrac{4}{3}$.

Substituting values in cosec² y - cot² y

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

Hence, cosec² y - cot² y = 1.

(iv) In right-angled ∆ACD,

sin x = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{CD}{AC} = \dfrac{5}{13}$.

In right-angled ∆ABD,

sin y = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{AD}{AB} = \dfrac{12}{20} = \dfrac{3}{5}$.

Substituting values in $\dfrac{5}{\text{sin x}} + \dfrac{3}{\text{sin y}} - 3\text{ cot y}$, we get :

$= \dfrac{5}{\dfrac{5}{13}} + \dfrac{3}{\dfrac{3}{5}} - 3 \times \dfrac{4}{3} \\[1em] = 13 + 5 - 4 \\[1em] = 14.$

Hence, $\dfrac{5}{\text{sin x}} + \dfrac{3}{\text{sin y}} - 3\text{ cot y} = 14$