In the given figure, ∠B = 90° and ∠ADB = x°. Find :

(i) sin ∠CAB

(ii) cos² C° + sin² C°

(iii) tan x° - cos x° + 3 sin x°

(i) In triangle CAB, using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AC² = AB² + BC²

⇒ 20² = 12² + BC²

⇒ 400 = 144 + BC²

⇒ BC² = 400 - 144

⇒ BC² = 256

⇒ BC = $\sqrt{256}$

⇒ BC = 16

sin ∠CAB = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

sin ∠CAB = $\dfrac{BC}{AC} = \dfrac{16}{20} = \dfrac{4}{5}$

Hence, sin ∠CAB = $\dfrac{4}{5}$

(ii) cos² C° + sin² C°

= $\Big(\dfrac{\text{Base}}{\text{Hypotenuse}}\Big)^2 + \Big(\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\Big)^2$

= $\Big(\dfrac{CB}{AC}\Big)^2 + \Big(\dfrac{AB}{AC}\Big)^2$

= $\dfrac{(CB)^2}{(AC)^2} + \dfrac{(AB)^2}{(AC)^2}$

= $\dfrac{(CB)^2 + (AB)^2}{(AC)^2}$

= $\dfrac{(AC)^2}{(AC)^2}$ (Using Pythagoras theorem)

= 1

Hence, cos² C° + sin² C° = 1.

(iii) In triangle ABD, using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AD² = AB² + BD²

⇒ 15² = 12² + BD²

⇒ 225 = 144 + BD²

⇒ BD² = 225 - 144

⇒ BD² = 81

⇒ BD = $\sqrt{81}$

⇒ BD = 9

Now, tan x° - cos x° + 3 sin x°

= $\dfrac{\text{Perpendicular}}{\text{Base}} - \dfrac{\text{Base}}{\text{Hypotenuse}} + 3 \times \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\dfrac{AB}{BD} - \dfrac{BD}{AD} + 3 \times \dfrac{AB}{AD}$

= $\dfrac{12}{9} - \dfrac{9}{15} + 3 \times \dfrac{12}{15}$

= $\dfrac{4}{3} - \dfrac{3}{5} + \dfrac{12}{5}$

= $\dfrac{20 - 9 + 36}{15}$

= $\dfrac{47}{15}$

= $3\dfrac{2}{15}$

Hence, tan x° - cos x° + 3 sin x° = $3\dfrac{2}{15}$.