From the figure (1) given below, find the values of :

(i) 2 sin y - cos y

(ii) 2 sin x - cos x

(iii) 1 - sin x + cos y

(iv) 2 cos x - 3 sin y + 4 tan x

In right-angled ∆BCD,

Using pythagoras theorem we get,

⇒ BC² = BD² + CD²

⇒ BC² = 9² + 12²

⇒ BC² = 81 + 144

⇒ BC² = 225

⇒ BC = $\sqrt{225}$

⇒ BC = 15.

In a right-angled ∆ABC,

Using pythagoras theorem

⇒ AC² = AB² + BC²

⇒ AB² = AC² - BC²

⇒ AB² = 25² - 15²

⇒ AB² = 625 - 225 = 400

⇒ AB = $\sqrt{400}$

⇒ AB = 20

(i) We know that

In right-angled ∆BCD,

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

= $\dfrac{BD}{BC} = \dfrac{9}{15} = \dfrac{3}{5}.$

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

= $\dfrac{CD}{BC} = \dfrac{12}{15} = \dfrac{4}{5}.$

Substituting values in 2 sin y - cos y we get :

$\Rightarrow \text{2 sin y - cos y} = 2 \times \dfrac{3}{5} - \dfrac{4}{5} \\[1em] = \dfrac{6}{5} - \dfrac{4}{5} \\[1em] = \dfrac{2}{5}.$

Hence, 2 sin y - cos y = $\dfrac{2}{5}$.

(ii) In right-angled ∆ABC

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

= $\dfrac{BC}{AC} = \dfrac{15}{25} = \dfrac{3}{5}.$

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

= $\dfrac{AB}{AC} = \dfrac{20}{25} = \dfrac{4}{5}.$

Substituting the values in 2 sin x - cos x we get :

$\Rightarrow \text{2 sin x - cos x} = 2 \times \dfrac{3}{5} - \dfrac{4}{5} \\[1em] = \dfrac{6}{5} - \dfrac{4}{5} \\[1em] = \dfrac{2}{5}.$

Hence, 2 sin x - cos x = $\dfrac{2}{5}$.

(iii) Substituting values we get :

$\Rightarrow \text{1 - sin x + cos y} = 1 - \dfrac{3}{5} + \dfrac{4}{5} \\[1em] = \dfrac{5 - 3 + 4}{5} \\[1em] = \dfrac{6}{5}.$

Hence, 1 - sin x + cos y = $\dfrac{6}{5}$.

(iv) By formula,

tan x = $\dfrac{\text{sin x}}{\text{cos x}} = \dfrac{\dfrac{3}{5}}{\dfrac{4}{5}} = \dfrac{3}{4}$.

Substituting values we get :

$\Rightarrow \text{2 cos x - 3 sin y + 4 tan x} = 2 \times \dfrac{4}{5} - 3 \times \dfrac{3}{5} + 4 \times \dfrac{3}{4}\\[1em] = \dfrac{8}{5} - \dfrac{9}{5} + 3 \\[1em] = -\dfrac{8 - 9 + 15}{5} \\[1em] = \dfrac{14}{5}.$

Hence, 2 cos x - 3 sin y + 4 tan x = $\dfrac{14}{5}$.