Trigonometrical Ratios

In right-angled triangle OMP,

By Pythagoras theorem, we get

⇒ OP² = OM² + MP²

⇒ MP² = OP² - OM²

⇒ MP² = (15)² - (12)²

⇒ MP² = 225 - 144

⇒ MP² = 81

⇒ MP = $\sqrt{81}$ = 9.

(i) sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{MP}{OP} = \frac{9}{15} = \frac{3}{5}$.

Hence, sin θ = $\frac{3}{5}$.

(ii) cos θ = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{OM}{OP} = \frac{12}{15} = \frac{4}{5}$.

Hence, cos θ = $\frac{4}{5}$.

(iii) tan θ = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{MP}{OM} = \frac{9}{12} = \frac{3}{4}$.

Hence, tan θ = $\frac{3}{4}$.

(iv) cot θ = $\frac{\text{Base}}{\text{Perpendicular}}$

= $\frac{OM}{MP} = \frac{12}{9} = \frac{4}{3}$.

Hence, cot θ = $\frac{4}{3}$.

(v) sec θ = $\frac{\text{Hypotenuse}}{\text{Base}}$

= $\frac{OP}{OM} = \frac{15}{12} = \frac{5}{4}$.

Hence, sec θ = $\frac{5}{4}$.

(vi) cosec θ = $\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

= $\frac{OP}{MP} = \frac{15}{9} = \frac{5}{3}$.

Hence, cosec θ = $\frac{5}{3}$.

In right-angled triangle ABC,

By Pythagoras theorem, we get

⇒ AB² = AC² + BC²

⇒ AB² = (12)² + (5)²

⇒ AB² = 144 + 25

⇒ AB² = 169

⇒ AB = $\sqrt{169}$ = 13

(i) sin A = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{BC}{AB} = \frac{5}{13}$.

Hence, sin A = $\frac{5}{13}$.

(ii) cos A = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AC}{AB} = \frac{12}{13}$.

Hence, cos A = $\frac{12}{13}$.

(iii) Substituting values of sin A and cos A in sin² A + cos² A :

⇒ sin² A + cos² A = $\Big(\frac{5}{13}\Big)^2 + \Big(\frac{12}{13}\Big)^2$

= $\frac{25}{169} + \frac{144}{169}$

= $\frac{169}{169}$ = 1.

Hence, sin² A + cos² A = 1.

(iv) sec² A - tan² A = $\Big(\frac{AB}{AC}\Big)^2 - \Big(\frac{BC}{AC}\Big)^2$

= $\Big(\frac{13}{12}\Big)^2 - \Big(\frac{5}{12}\Big)^2$

= $\frac{169}{144} - \frac{25}{144}$

= $\frac{144}{144}$ = 1.

Hence, sec² A - tan² A = 1.

In right-angled triangle ABC,

By Pythagoras theorem, we get

⇒ BC² = AC² + AB²

⇒ AC² = BC² - AB²

⇒ AC² = (10)² - (6)²

⇒ AC² = 100 - 36

⇒ AC² = 64

⇒ AC = $\sqrt{64}$

⇒ AC = 8.

(i) sin B = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{AC}{BC} = \frac{8}{10} = \frac{4}{5}$.

Hence, sin B = $\frac{4}{5}$.

(ii) cos C = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AC}{BC} = \frac{8}{10} = \frac{4}{5}.$

Hence, cos C = $\frac{4}{5}$.

(iii) sin C = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{AB}{BC} = \frac{6}{10} = \frac{3}{5}$.

Substituting values of sin B and sin C in sin B + sin C we get :

⇒ sin B + sin C = $\frac{4}{5} + \frac{3}{5} = \frac{7}{5}$.

Hence, sin B + sin C = $\frac{7}{5}$.

(iv) sin B = $\frac{4}{5}$, sin C = $\frac{3}{5}$, cos C = $\frac{4}{5}$.

cos B = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AB}{BC} = \frac{6}{10} = \frac{3}{5}$.

Substituting values in equation sin B cos C + sin C cos B we get :

$$\begin{aligned}= \frac{4}{5} \times \frac{4}{5} + \frac{3}{5} \times \frac{3}{5} \\[1em] = \frac{16}{25} + \frac{9}{25} \\[1em] = \frac{25}{25} = 1.\end{aligned}$$

Hence, sin B cos C + sin C cos B = 1.

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 = $\frac{\text{Perpendicular}}{\text{Base}}$

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

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

(ii) In right-angled ∆ABD,

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

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

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

(iii) In right-angled ∆ABD,

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

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

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

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

Substituting values in cosec² y - cot² y

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

Hence, cosec² y - cot² y = 1.

(iv) In right-angled ∆ACD,

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

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

In right-angled ∆ABD,

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

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

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

$$\begin{aligned}= \frac{5}{\frac{5}{13}} + \frac{3}{\frac{3}{5}} - 3 \times \frac{4}{3} \\[1em] = 13 + 5 - 4 \\[1em] = 14.\end{aligned}$$

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

From figure,

⇒ BD = BC - CD = 21 - 5 = 16.

In △ADC,

⇒ AC² = AD² + DC² [By pythagoras theorem]

⇒ AD² = AC² - DC²

⇒ AD² = 13² - 5²

⇒ AD² = 169 - 25

⇒ AD² = 144

⇒ AD = $\sqrt{144}$

⇒ AD = 12.

In △ABD,

⇒ AB² = AD² + BD² [By pythagoras theorem]

⇒ AB² = 12² + 16²

⇒ AB² = 144 + 256

⇒ AB² = 400

⇒ AB = $\sqrt{400}$

⇒ AB = 20.

By formula,

sec θ = $\frac{\text{Hypotenuse}}{\text{Base}}$

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

Hence, sec θ = $\frac{5}{4}$.

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 = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

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

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

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

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

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

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

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

Substituting values in cot² x - cosec² x we get :

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

Hence, cot² x - cosec² x = -1.

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

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

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

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

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

Given,

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

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

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 = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

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

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

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

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

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

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

(ii) In right-angled ∆ABC

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

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

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

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

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

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

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

(iii) Substituting values we get :

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

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

(iv) By formula,

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

Substituting values we get :

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

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

Given, ΔABC is a right angled at B.

sin θ = $\frac{5}{13}$ and BC = 24 cm

$$\begin{aligned}\Rightarrow \text{sin θ} = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\\[1em] \Rightarrow \frac{5}{13} = \frac{AB}{AC}\end{aligned}$$

Let AB = 5k and AC = 13k.

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ (13k)² = (5k)² + 24²

⇒ 169k² = 25k² + 576

⇒ 169k² - 25k² = 576

⇒ 144k² = 576

⇒ k² = $\frac{576}{144}$

⇒ k² = 4

⇒ k = $\sqrt{4}$

⇒ k = ± 2

Since, length cannot be negative.

⇒ k = 2

So, AB = 5k = 5 x 2 = 10 cm,

and AC = 13k = 13 x 2 = 26 cm.

Hence, AB = 10 cm and AC = 26 cm.

Let ABC be a right angled triangle with ∠B = 90°.

Given,

tan A = $\frac{5}{12}$.

By formula,

tan A = $\frac{\text{Perpendicular}}{\text{Base}} = \frac{BC}{AB} = \frac{5}{12}$

Let BC = 5x and AB = 12x.

From right-angled ∆ABC

By pythagoras theorem, we get

⇒ AC² = BC² + AB²

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169x²

⇒ AC = $\sqrt{169x^2}$

⇒ AC = 13x.

(i) By formula,

cos A = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AB}{AC} = \frac{12x}{13x} = \frac{12}{13}$.

Hence, cos A = $\frac{12}{13}$.

(ii) By formula,

cosec A = $\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

= $\frac{AC}{BC} = \frac{13x}{5x} = \frac{13}{5}$.

cot A = $\frac{\text{Base}}{\text{Perpendicular}}$

= $\frac{AB}{BC}$ = $\frac{12x}{5x}$ = $\frac{12}{5}$.

Substituting values in cosec A - cot A, we get :

$$\begin{aligned}\Rightarrow \text{cosec A - cot A} = \frac{13}{5} - \frac{12}{5} \\[1em] = \frac{1}{5}.\end{aligned}$$

Hence, cosec A - cot A = $\frac{1}{5}$.

(a) In right ∆ABC

∠A = 90°

AB = 7 cm

BC - AC = 1 cm

⇒ BC = 1 + AC

We know that,

⇒ BC² = AB² + AC² [By pythagoras theorem]

⇒ (1 + AC)² = AB² + AC²

⇒ 1 + AC² + 2AC = 7² + AC²

⇒ 1 + AC² + 2AC = 49 + AC²

⇒ 2AC = 49 - 1

⇒ 2AC = 48

⇒ AC = $\frac{48}{2}$ = 24 cm.

∴ BC = 1 + AC = 25 cm.

(i) sin C = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{AB}{BC} = \frac{7}{25}$

Hence, sin C = $\frac{7}{25}$.

(ii) tan B = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{AC}{AB} = \frac{24}{7}$

Hence, tan B = $\frac{24}{7}$.

In right ∆PQR,

∠Q = 90°

PQ = 40 cm

Given,

⇒ PR + QR = 50 cm

⇒ PR = 50 - QR

Using Pythagoras theorem we get,

⇒ PR² = PQ² + QR²

⇒ (50 - QR)² = (40)² + QR²

⇒ 50² + QR² - 100QR = 1600 + QR²

⇒ QR² - QR² - 100QR = 1600 - 2500

⇒ -100QR = -900

⇒ 100QR = 900

⇒ QR = $\frac{900}{100}$ = 9.

∴ PR = 50 - QR = 50 - 9 = 41.

(i) sin P = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{QR}{PR} = \frac{9}{41}$.

Hence, sin P = $\frac{9}{41}$.

(ii) cos P = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{PQ}{PR} = \frac{40}{41}$.

Hence, cos P = $\frac{40}{41}$.

(iii) tan R = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{PQ}{QR} = \frac{40}{9}$.

Hence, tan R = $\frac{40}{9}$.

Draw AD perpendicular to BC.

D is mid-point of BC [∵ Perpendicular drawn to the unequal side of an isosceles triangle from the apex vertex bisects the side]

∴ BD = DC = 9 cm.

In right-angled triangle ABD,

⇒ AB² = AD² + BD²

⇒ 15² = AD² + 9²

⇒ AD² = (15)² - (9)²

⇒ AD² = 225 - 81

⇒ AD² = 144

⇒ AD = $\sqrt{144}$ = 12 cm.

(i) cos ∠ABC = $\frac{BD}{AB}$

= $\frac{9}{15} = \frac{3}{5}$.

Hence, cos ∠ABC = $\frac{3}{5}$.

(ii) sin ∠ACB = $\frac{AD}{AC}$

= $\frac{12}{15} = \frac{4}{5}$.

Hence, sin ∠ACB = $\frac{4}{5}$.

Draw AD perpendicular to BC.

D is the mid point of BC [∵ Perpendicular drawn to the unequal side of an isosceles triangle from the apex vertex bisects the side]

So, BD = CD = 3 cm.

In right-angled ∆ABD,

Using pythagoras theorem we get :

⇒ AB² = AD² + BD²

⇒ AD² = AB² - BD²

⇒ AD² = 5² - 3²

⇒ AD² = 25 - 9

⇒ AD² = 16

⇒ AD = $\sqrt{16}$

⇒ AD = 4 cm.

(i) In right-angled ∆ACD,

sin C = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{AD}{AC} = \frac{4}{5}$.

Hence, sin C = $\frac{4}{5}$.

(ii) In right-angled ∆ABD,

tan B = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{AD}{BD} = \frac{4}{3}$.

Hence, tan B = $\frac{4}{3}$.

(iii) In right-angled ∆ACD,

tan C = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{AD}{CD} = \frac{4}{3}$.

In right-angled ∆ABD,

cot B = $\frac{\text{base}}{\text{Perpendicular}}$

= $\frac{BD}{AD} = \frac{3}{4}$.

Substituting values in tan C - cot B we get :

$$\begin{aligned}\text{tan C} - \text{cot B} = \frac{4}{3} - \frac{3}{4} \\[1em] = \frac{16 - 9}{12} \\[1em] = \frac{7}{12}.\end{aligned}$$

Hence, tan C - cot B = $\frac{7}{12}$.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (2)² + (1)²

⇒ AC² = 4 + 1 = 5

⇒ AC = $\sqrt{5}$.

Calculating sin θ, we get :

$$\begin{aligned}\text{sin θ} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \frac{AB}{AC} \\[1em] = \frac{2}{\sqrt{5}}.\end{aligned}$$

Calculating tan θ, we get :

$$\begin{aligned}\text{tan θ} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] = \frac{AB}{BC} \\[1em] = \frac{2}{1} = 2.\end{aligned}$$

Substituting value of sin θ and tan θ in sin² θ + tan² θ, we get :

$$\begin{aligned}\text{sin}^2 \text{ θ} + \text{tan}^2 \text{ θ} = \Big(\frac{2}{\sqrt{5}}\Big)^2 + 2^2 \\[1em] = \frac{4}{5} + 4 \\[1em] = \frac{4 + 20}{5} \\[1em] = \frac{24}{5} = 4\frac{4}{5}.\end{aligned}$$

Hence, sin² θ + tan² θ = $4\frac{4}{5}$.

(i) From figure,

tan C = $\frac{\text{Perpendicular}}{\text{Base}}$ = $\frac{AD}{DC}$

⇒ 1 = $\frac{AD}{DC}$

⇒ AD = DC.

sin B = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$ = $\frac{AD}{AB}$

⇒ $\frac{4}{5}$ = $\frac{AD}{AB}$

Let AD = 4k and AB = 5k.

In right angle triangle ABD,

⇒ AB² = AD² + BD²

⇒ (5k)² = (4k)² + 15²

⇒ 25k² = 16k² + 225

⇒ 9k² = 225

⇒ k² = $\frac{225}{9}$ = 25

⇒ k = $\sqrt{25}$ = 5.

AD = 4k = 4 × 5 = 20

AB = 5k = 5 × 5 = 25.

DC = AD = 20.

In right angle triangle ADC,

⇒ AC² = AD² + DC²

⇒ AC² = 20² + 20²

⇒ AC² = 400 +400

⇒ AC² = 800

⇒ AC = $\sqrt{800} = 20\sqrt{2}$.

Hence, AD = 20, AB = 25, DC = 20 and AC = $20\sqrt{2}$.

(ii) Calculating tan B, we get :

$$\begin{aligned}\text{tan B} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] = \frac{AD}{BD} \\[1em] = \frac{20}{15} = \frac{4}{3}.\end{aligned}$$

Calculating cos B, we get :

$$\begin{aligned}\text{cos B} = \frac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \frac{BD}{AB} \\[1em] = \frac{15}{25} = \frac{3}{5}.\end{aligned}$$

Substituting value of tan B and cos B in L.H.S. of the equation, $\text{tan}^2 \text{ B} - \frac{1}{\text{cos}^2 \text{ B}}$ = -1, we get :

$$\begin{aligned}\Rightarrow \text{tan}^2 B - \frac{1}{\text{cos}^2 B} = \Big(\frac{4}{3}\Big)^2 - \frac{1}{\Big(\frac{3}{5}\Big)^2} \\[1em] = \frac{16}{9} - \frac{1}{\frac{9}{25}} \\[1em] = \frac{16}{9} - \frac{25}{9} \\[1em] = \frac{-9}{9} \\[1em] = -1.\end{aligned}$$

Hence, proved that tan² B - $\frac{1}{\text{cos}^2 B}$ = -1.

(i) By formula,

⇒ sin² θ + cos² θ = 1

Substituting values we get,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{5}\Big)^2 + \text{cos}^2 \text{ θ} = 1 \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = 1 - \frac{9}{25} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \frac{25 - 9}{25} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \frac{16}{25} \\[1em] \Rightarrow \text{cos θ} = \sqrt{\frac{16}{25}} \\[1em] \Rightarrow \text{cos θ} = \pm \frac{4}{5}.\end{aligned}$$

Since, θ is an acute angle and value of cos is positive in first quadrant.

∴ cos θ = $\frac{4}{5}$.

Hence, cos θ = $\frac{4}{5}$.

(ii) By formula,

tan θ = $\frac{\text{sin θ}}{\text{cos θ}}$

= $\frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$.

Hence, tan θ = $\frac{3}{4}$.

By formula,

sec² θ = 1 + tan² θ

Substituting values we get,

$$\begin{aligned}\Rightarrow \text{sec}^2 \text{ θ} = 1 + \Big(\frac{5}{12}\Big)^2 \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = 1 + \frac{25}{144} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \frac{144 + 25}{144} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \frac{169}{144} \\[1em] \Rightarrow \text{sec θ} = \sqrt{\frac{169}{144}} \\[1em] \Rightarrow \text{sec θ} = \pm \frac{13}{12}.\end{aligned}$$

Since, θ is an acute angle and value of sec is positive in first quadrant.

∴ sec θ = $\frac{13}{12}$.

By formula,

cos θ = $\frac{1}{\text{sec θ}} = \frac{1}{\frac{13}{12}} = \frac{12}{13}.$

sin² θ + cos² θ = 1

Substituting values we get,

$$\begin{aligned}\Rightarrow \text{sin}^2 \text{ θ} + \Big(\frac{12}{13}\Big)^2 = 1 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = 1 - \Big(\frac{12}{13}\Big)^2\\[1em] \Rightarrow \text{sin}^2 \text{ θ} = 1 - \frac{144}{169} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \frac{169 - 144}{169} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \frac{25}{169} \\[1em] \Rightarrow \text{sin θ} = \sqrt{\frac{25}{169}} \\[1em] \Rightarrow \text{sin θ} = \pm \frac{5}{13}.\end{aligned}$$

Since, θ is an acute angle and value of sin is positive in first quadrant.

∴ sin θ = $\frac{5}{13}$.

Hence, sin θ = $\frac{5}{13}$ and cos θ = $\frac{12}{13}$.

Given,

$$\begin{aligned}\Rightarrow \text{sin θ} = \frac{6}{10} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \frac{36}{100} \\[1em] \Rightarrow 1 - \text{cos}^2 \text{ θ} = \frac{36}{100} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = 1 - \frac{36}{100} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \frac{100 - 36}{100} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \frac{64}{100} \\[1em] \Rightarrow \text{cos θ} = \sqrt{\frac{64}{100}} \\[1em] \Rightarrow \text{cos θ} = \frac{8}{10} \\[1.5em] \text{tan θ} = \frac{\text{sin θ}}{\text{cos θ}} \\[1em] = \frac{\frac{6}{10}}{\frac{8}{10}} = \frac{6}{8}.\end{aligned}$$

Substituting values in cos θ + tan θ we get,

$$\begin{aligned}\text{cos θ + tan θ }= \frac{8}{10} + \frac{6}{8} \\[1em] = \frac{32 + 30}{40} \\[1em] = \frac{62}{40} \\[1em] = \frac{31}{20} \\[1em] = 1\frac{11}{20}\end{aligned}$$

Hence, cos θ + tan θ = $1\frac{11}{20}.$

Given,

$$\begin{aligned}\phantom{\Rightarrow} \text{tan θ} = \frac{4}{3} \\[1em] \Rightarrow \text{tan}^2 \text{ θ} = \frac{16}{9} \\[1em] \phantom{\Rightarrow} \text{sec}^2 \text{ θ} = 1 + \text{tan}^2 θ \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = 1 + \frac{16}{9} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \frac{9 + 16}{9} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} = \frac{25}{9} \\[1em] \Rightarrow \frac{1}{\text{cos}^2 \text{ θ}} = \frac{25}{9} \\[1em] \Rightarrow \text{cos}^2 \text{ θ} = \frac{9}{25} \\[1em] \Rightarrow \text{cos θ} = \sqrt{\frac{9}{25}} \\[1em] \Rightarrow \text{cos θ} = \pm\frac{3}{5} \\[1em] \Rightarrow \text{cos θ} = \frac{3}{5} [\text{Given, cos θ is positive.}].\end{aligned}$$

By formula,

sin² θ + cos² θ = 1

Substituting values in sin² θ + cos² θ = 1 we get :

$$\begin{aligned}\Rightarrow \text{sin}^2 \text{ θ} + \Big(\frac{3}{5}\Big)^2 = 1 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = 1 - \frac{9}{25} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \frac{25 - 9}{25} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} = \frac{16}{25} \\[1em] \Rightarrow \text{sin θ} = \sqrt{\frac{16}{25}} \\[1em] \Rightarrow \text{sin θ} = \pm\frac{4}{5} \\[1em] \Rightarrow \text{sin θ} = \frac{4}{5} \space [\text{Given, sin θ is positive.}].\end{aligned}$$

Substituting values in sin θ + cos θ we get :

sin θ + cos θ = $\frac{4}{5} + \frac{3}{5} = \frac{7}{5} = 1\frac{2}{5}.$

Hence, sin θ + cos θ = $1\frac{2}{5}$.

Let ABC be a right angle triangle with ∠C = θ and ∠B = 90°.

By formula,

cosec θ = $\frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{AC}{AB}$.

Substituting values we get :

$\Rightarrow \frac{\sqrt{5}}{1} = \frac{AC}{AB}$

Let AC = $\sqrt{5}$k and AB = k.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ ($\sqrt{5}$k)² = BC² + k²

⇒ 5k² = BC² + k²

⇒ BC² = 5k² - k²

⇒ BC² = 4k²

⇒ BC = $\sqrt{4\text{k}^2}$

⇒ BC = 2k.

cot θ = $\frac{\text{Base}}{\text{Perpendicular}}$

= $\frac{BC}{AB}$ = $\frac{2k}{k}$ = $\frac{2}{1}$

cos θ = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{BC}{AC}$ = $\frac{2k}{\sqrt{5}k}$ = $\frac{2}{\sqrt{5}}$

$$\begin{aligned}\therefore \text{cot θ - cos θ} = \frac{2}{1} - \frac{2}{\sqrt{5}} \\[1em] = \frac{2\sqrt{5} - 2}{\sqrt{5}} \\[1em] = \frac{2(\sqrt{5} - 1)}{\sqrt{5}}.\end{aligned}$$

Hence, cot θ - cos θ = $\frac{2(\sqrt{5} - 1)}{\sqrt{5}}.$

Let ABC be a right angle triangle with ∠C = θ and ∠B = 90°.

By formula,

sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{AC}$ ........(1)

Given,

sin θ = $\frac{p}{q}$ .........(2)

From (1) and (2) we get,

$\frac{AB}{AC} = \frac{p}{q}$

Let AB = px and AC = qx.

In right angled triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ BC² = AC² - AB²

⇒ BC² = (qx)² - (px)²

⇒ BC² = q²x² - p²x²

⇒ BC² = x²(q² - p²)

⇒ BC = $\sqrt{x^2(q^2 - p^2)}$

⇒ BC = $x\sqrt{(q^2 - p^2)}$

In right angled triangle ABC,

By formula,

cos θ = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{BC}{AC} = \frac{x\sqrt{q^2 - p^2}}{qx} = \frac{\sqrt{q^2 - p^2}}{q}$.

Substituting values in cos θ + sin θ we get,

$$\begin{aligned}\Rightarrow \text{cos θ + sin θ} = \frac{\sqrt{q^2 - p^2}}{q} + \frac{p}{q} \\[1em] = \frac{p + \sqrt{q^2 - p^2}}{q}.\end{aligned}$$

Hence, cos θ + sin θ = $\frac{p + \sqrt{q^2 - p^2}}{q}.$

Given,

tan θ = $\frac{8}{15}$ ..........(1)

Let ABC be a right angle triangle with ∠C = θ and ∠B = 90°.

By formula,

tan θ = $\frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$ .........(2)

From (1) and (2) we get,

$\frac{AB}{BC} = \frac{8}{15}$

Let AB = 8x and BC = 15x.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (8x)² + (15x)²

⇒ AC² = 64x² + 225x²

⇒ AC² = 289x²

⇒ AC = $\sqrt{289x^2}$

⇒ AC = 17x.

By formula,

sec θ = $\frac{\text{Hypotenuse}}{\text{Base}}$

= $\frac{AC}{BC} = \frac{17x}{15x} = \frac{17}{15}$.

cosec θ = $\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

= $\frac{AC}{AB} = \frac{17x}{8x} = \frac{17}{8}$.

Substituting value in sec θ + cosec θ we get :

$$\begin{aligned}\Rightarrow \text{sec θ + cosec θ} = \frac{17}{15} + \frac{17}{8} \\[1em] = \frac{136 + 255}{120} \\[1em] = \frac{391}{120} \\[1em] = 3\frac{31}{120}.\end{aligned}$$

Hence, sec θ + cosec θ = $3\frac{31}{120}$.

Let triangle ABC be a right-angled triangle at B and A is an acute angle.

Given,

⇒ 13 sin A = 5

⇒ sin A = $\frac{5}{13}$

⇒ $\frac{BC}{AC} = \frac{5}{13}$

Let BC = 5k and AC = 13k.

In right angled triangle ABC,

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AB² = AC² - BC²

⇒ AB² = (13k)² - (5k)²

⇒ AB² = 169k² - 25k²

⇒ AB² = 144k²

⇒ AB = $\sqrt{144k^2}$

⇒ AB = 12k.

By formula,

cos A = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AB}{AC} = \frac{12k}{13k} = \frac{12}{13}$.

tan A = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{BC}{AB} = \frac{5k}{12k} = \frac{5}{12}$.

Substituting values we get,

$$\begin{aligned}\phantom{=} \frac{\text{5 sin A - 2 cos A}}{\text{tan A}} \\[1em] = \frac{5 \times \frac{5}{13} - 2 \times \frac{12}{13}}{\frac{5}{12}} \\[1em] = \frac{\frac{25}{13} - \frac{24}{13}}{\frac{5}{12}} \\[1em] = \frac{\frac{1}{13}}{\frac{5}{12}} \\[1em] = \frac{12}{65}.\end{aligned}$$

Hence, $\frac{\text{5 sin A - 2 cos A}}{\text{tan A}} = \frac{12}{65}$.

Let triangle ABC be a right-angled at B and A is a acute angle.

Given,

cosec A = $\sqrt{2}$

⇒ cosec A = $\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

= $\frac{AC}{BC} = \frac{\sqrt{2}}{1}$

Let AC = $\sqrt{2}$k and BC = k.

In right angled triangle ABC,

By using pythagoras theorem,

AC² = AB² + BC²

($\sqrt{2}k$)² = AB² + k²

2k² = AB² + k²

AB² = 2k² - k²

AB² = k²

AB = $\sqrt{\text{k}}$ = k.

By formula,

sin A = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{BC}{AC} = \frac{k}{\sqrt{2}k} = \frac{1}{\sqrt{2}}$.

tan A = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{BC}{AB} = \frac{k}{k} = 1$.

cot A = $\frac{1}{\text{tan A}} = \frac{1}{1} = 1$.

cos A = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AB}{AC} = \frac{k}{\sqrt{2}k} = \frac{1}{\sqrt{2}}$.

Substituting these values in $\frac{\text{2 sin}^2 A + 3\text{ cot}^2 A}{\text{tan}^2 A - \text{cos}^2 A}$ we get,

$$\begin{aligned}\Rightarrow \frac{2 \times \Big(\frac{1}{\sqrt{2}}\Big)^2 + 3 \times 1^2}{1^2 - \Big(\frac{1}{\sqrt{2}}\Big)^2} \\[1em] \Rightarrow \frac{2 \times \frac{1}{2} + 3 \times 1}{1 - \frac{1}{2}} \\[1em] \Rightarrow \frac{1 + 3}{\frac{1}{2}} \\[1em] \Rightarrow 4 \times 2 \\[1em] \Rightarrow 8.\end{aligned}$$

Hence, $\frac{\text{2 sin}^2 A + 3\text{ cot}^2 A}{\text{tan}^2 A - \text{cos}^2 A}$ = 8.

Since, diagonals of rhombus bisect each other.

∴ O is the mid point of AC.

In right angled ∆COD,

⇒ CD² = OC² + OD² [By pythagoras theorem]

⇒ CD² = 4² + 3²

⇒ CD² = 16 + 9

⇒ CD² = 25

⇒ CD = $\sqrt{25}$

⇒ CD = 5 cm.

sin ∠OCD = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

sin ∠OCD = $\frac{OD}{CD} = \frac{3}{5}$.

Hence, sin ∠OCD = $\frac{3}{5}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{(cos θ + sin θ)}}{\text{(cos θ - sin θ)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \frac{\frac{\text{(cos θ + sin θ)}}{\text{cos θ}}}{\frac{\text{(cos θ - sin θ)}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{\frac{\text{cos θ}}{\text{cos θ}} + \frac{\text{sin θ}}{\text{cos θ}}}{\frac{\text{cos θ}}{\text{cos θ}} - \frac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{1 + \text{tan θ}}{1 - \text{tan θ}} [\text{As tan θ} = \frac{\text{sin θ}}{\text{cos θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \frac{1 + \frac{5}{12}}{1 - \frac{5}{12}} \\[1em] \Rightarrow \frac{\frac{12 + 5}{12}}{\frac{12 - 5}{12}} \\[1em] \Rightarrow \frac{\frac{17}{12}}{\frac{7}{12}} \\[1em] \Rightarrow \frac{17}{7} = 2\frac{3}{7}.\end{aligned}$$

Hence, $\frac{\text{(cos θ + sin θ)}}{\text{(cos θ - sin θ)}} = 2\frac{3}{7}$.

Given,

⇒ 5 cos A - 12 sin A = 0

⇒ 5 cos A = 12 sin A

⇒ $\frac{\text{sin A}}{\text{cos A}} = \frac{5}{12}$

⇒ tan A = $\frac{5}{12}$.

We need to find the value of

$$\begin{aligned}\frac{\text{(sin A + cos A)}}{\text{(2cos A - sin A)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \frac{\frac{\text{(sin A + cos A)}}{\text{cos A}}}{\frac{\text{(2cos A - sin A)}}{\text{cos A}}} \\[1em] \Rightarrow \frac{\frac{\text{sin A}}{\text{cos A}} + \frac{\text{cos A}}{\text{cos A}}}{\frac{\text{2cos A}}{\text{cos A}} - \frac{\text{sin A}}{\text{cos A}}} \\[1em] \Rightarrow \frac{\text{tan A + 1}}{{2 - \text{tan A}}} \space [\because \text{tan A} = \frac{\text{sin A}}{\text{cos A}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \frac{\frac{5}{12} + 1}{2 - \frac{5}{12}} \\[1em] \Rightarrow \frac{\frac{5 + 12}{12}}{\frac{24 - 5}{12}} \\[1em] \Rightarrow \frac{\frac{17}{12}}{\frac{19}{12}} \\[1em] \Rightarrow \frac{17 \times 12}{12 \times 19} \\[1em] \Rightarrow \frac{17}{19}.\end{aligned}$$

Hence, $\frac{\text{sin A + cos A}}{\text{2 cos A - sin A}} = \frac{17}{19}$.

Given,

$$\begin{aligned}\frac{(\text{p sin θ - q cos θ})}{(\text{p sin θ + q cos θ)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \frac{\frac{(\text{p sin θ - q cos θ})}{\text{cos θ}}}{\frac{(\text{p sin θ + q cos θ)}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{\frac{\text{p sin θ}}{\text{cos θ}} - \frac{\text{q cos θ}}{\text{cos θ}}}{\frac{\text{p sin θ}}{\text{cos θ}} + \frac{\text{q cos θ}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{\text{p tan θ - q}}{{\text{p tan θ + q}}} \space [\because \text{tan θ} = \frac{\text{sin θ}}{\text{cos θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \frac{p \times \frac{p}{q} - q}{p \times \frac{p}{q} + q} \\[1em] \Rightarrow \frac{\frac{p^2}{q} - q}{\frac{p^2}{q} + q} \\[1em] \Rightarrow \frac{\frac{p^2 - q^2}{q}}{\frac{p^2 + q^2}{q}} \\[1em] \Rightarrow \frac{p^2 - q^2}{p^2 + q^2}.\end{aligned}$$

Hence, $\frac{(\text{p sin θ - q cos θ})}{(\text{p sin θ + q cos θ)}} = \frac{p^2 - q^2}{p^2 + q^2}.$

Given,

⇒ 3 cot θ = 4

⇒ cot θ = $\frac{4}{3}$

$$\begin{aligned}\frac{\text{5 sin θ - 3 cos θ}}{\text{5 sin θ + 3 cos θ}} \\[1em] [\text{Dividing numerator and denominator by sin θ}] \\[1em] \Rightarrow \frac{\frac{\text{(5 sin θ - 3 cos θ)}}{\text{sin θ}}}{\frac{\text{(5 sin θ + 3 cos θ)}}{\text{sin θ}}} \\[1em] \Rightarrow \frac{\frac{\text{5 sin θ}}{\text{sin θ}} - \frac{\text{3 cos θ}}{\text{sin θ}}}{\frac{\text{5 sin θ}}{\text{sin θ}} + \frac{\text{3 cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \frac{\text{5 - 3 cot θ}}{{5 + \text{3 cot θ}}} \space [\because \text{cot θ} = \frac{\text{cos θ}}{\text{sin θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \frac{5 - 3 \times \frac{4}{3}}{5 + 3\times \frac{4}{3}} \\[1em] \Rightarrow \frac{5 - 4}{5 + 4} \\[1em] \Rightarrow \frac{1}{9}.\end{aligned}$$

Hence, $\frac{\text{5 sin θ - 3 cos θ}}{\text{5 sin θ + 3 cos θ}} = \frac{1}{9}$

(i) Given,

⇒ 5 cos θ - 12 sin θ = 0

⇒ 5 cos θ = 12 sin θ

⇒ $\frac{\text{sin θ}}{\text{cos θ}} = \frac{5}{12}$

⇒ tan θ = $\frac{5}{12}$.

$$\begin{aligned}\Rightarrow \frac{\text{(sin θ + cos θ)}}{\text{(2 cos θ - sin θ)}} \\[1em] [\text{Dividing numerator and denominator by cos θ}] \\[1em] \Rightarrow \frac{\frac{\text{(sin θ + cos θ)}}{\text{cos θ}}}{\frac{\text{(2 cos θ - sin θ)}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{\frac{\text{sin θ}}{\text{cos θ}} + \frac{\text{cos θ}}{\text{cos θ}}}{\frac{\text{2 cos θ}}{\text{cos θ}} - \frac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{\text{tan θ} + 1}{2 - \text{tan θ}} \space [\because \text{tan θ} = \frac{\text{sin θ}}{\text{cos θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \frac{\frac{5}{12} + 1}{2 - \frac{5}{12}} \\[1em] \Rightarrow \frac{\frac{5 + 12}{12}}{\frac{24 - 5}{12}} \Rightarrow \frac{\frac{17}{12}}{\frac{19}{12}} \\[1em] \Rightarrow \frac{17}{19}.\end{aligned}$$

Hence, $\frac{\text{(sin θ + cos θ)}}{\text{(2 cos θ - sin θ)}} = \frac{17}{19}$.

Given,

cosec θ = $\frac{13}{12}$

By formula,

⇒ cot² θ = cosec² θ - 1

$$\begin{aligned}\text{cot}^2 \text{ θ} = \Big(\frac{13}{12}\Big)^2 - 1 \\[1em] = \frac{169}{144} - 1 \\[1em] = \frac{169 - 144}{144} \\[1em] = \frac{25}{144} \\[1em] \Rightarrow \text{cot θ} = \sqrt{\frac{25}{144}} = \frac{5}{12}\end{aligned}$$

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} \\[1em] [\text{Dividing numerator and denominator by sin θ}] \\[1em] \Rightarrow \frac{\frac{\text{(2 sin θ - 3 cos θ)}}{\text{sin θ}}}{\frac{\text{(4 sin θ - 9 cos θ)}}{\text{sin θ}}} \\[1em] \Rightarrow \frac{\frac{\text{2 sin θ}}{\text{sin θ}} - \frac{\text{3 cos θ}}{\text{sin θ}}}{\frac{\text{4 sin θ}}{\text{sin θ}} - \frac{\text{9 cos θ}}{\text{sin θ}}} \\[1em] \Rightarrow \frac{\text{2 - 3 cot θ}}{{4 - \text{9 cot θ}}} \space [\because \text{cot θ} = \frac{\text{cos θ}}{\text{sin θ}}] \\[1em] \text{Substituting values we get} \\[1em] \Rightarrow \frac{2 - 3 \times \frac{5}{12}}{4 - 9\times \frac{5}{12}} \\[1em] \Rightarrow \frac{2 - \frac{5}{4}}{4 - \frac{15}{4}} \\[1em] \Rightarrow \frac{\frac{8 - 5}{4}}{\frac{16 - 15}{4}} \\[1em] \Rightarrow \frac{\frac{3}{4}}{\frac{1}{4}} \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, $\frac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} = 3.$

Given,

5 sin θ = 3

⇒ sin θ = $\frac{3}{5}$.

Solving,

$$\begin{aligned}\Rightarrow \frac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} \\[1em] \Rightarrow \frac{\frac{1}{\text{cos θ}} - \frac{\text{sin θ}}{\text{cos θ}}}{\frac{1}{\text{cos θ}} + \frac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{\frac{\text{1 - sin θ}}{\text{cos θ}}}{\frac{\text{1 + sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{\text{1 - sin θ}}{\text{1 + sin θ}} \\[1em] \Rightarrow \frac{1 - \frac{3}{5}}{1 + \frac{3}{5}} \\[1em] \Rightarrow \frac{\frac{5 - 3}{5}}{\frac{5 + 3}{5}} \\[1em] \Rightarrow \frac{\frac{2}{5}}{\frac{8}{5}} \\[1em] \Rightarrow \frac{2}{8} \\[1em] \Rightarrow \frac{1}{4}.\end{aligned}$$

Hence, $\frac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} = \frac{1}{4}$.

Given,

sin θ = cos θ

⇒ $\frac{\text{sin θ}}{\text{cos θ}} = 1$

⇒ tan θ = 1.

⇒ tan θ = tan 45°

⇒ θ = 45°.

⇒ sin 45° = cos 45° = $\frac{1}{\sqrt{2}}$

Substituting values in 2 tan² θ + sin² θ - 1 we get :

$$\begin{aligned}\Rightarrow 2(1)^2 + \Big(\frac{1}{\sqrt{2}}\Big)^2 - 1 \\[1em] \Rightarrow 2 + \frac{1}{2} - 1 \\[1em] \Rightarrow 1 + \frac{1}{2} \\[1em] \Rightarrow \frac{2 + 1}{2} \Rightarrow \frac{3}{2}.\end{aligned}$$

Hence, 2 tan² θ + sin² θ - 1 = $\frac{3}{2}$.

To prove,

cos θ tan θ = sin θ

We know that,

$\text{tan θ} = \frac{\text{sin θ}}{\text{cos θ}}$

Substituting value in L.H.S. of cos θ tan θ = sin θ we get,

$$\begin{aligned}\text{cos θ} \times \frac{\text{sin θ}}{\text{cos θ}} = \text{sin θ} \\[1em] \Rightarrow \text{sin θ} = \text{sin θ}\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence proved that cos θ tan θ = sin θ.

To prove,

sin θ cot θ = cos θ

We know that,

cot θ = $\frac{\text{cos θ}}{\text{sin θ}}$

Substituting value in L.H.S. of sin θ cot θ = cos θ we get,

$$\begin{aligned}\text{sin θ} \times \frac{\text{cos θ}}{\text{sin θ}} = \text{cos θ} \\[1em] \Rightarrow \text{cos θ} = \text{cos θ}\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence. proved that sin θ cot θ = cos θ.

To prove,

$\frac{\text{sin}^2 \text{ θ}}{\text{cos \text{ θ}}} + \text{cos \text{ θ}} = \frac{1}{\text{cos \text{ θ}}}$

Solving LHS of the above equation,

$$\begin{aligned}\phantom{=} \frac{\text{sin}^2 \text{ θ}}{\text{cos θ}} + \text{cos θ} \\[1em] = \frac{\text{sin}^2 \text{ θ} + \text{cos}^2 θ}{\text{cos θ}} \\[1em] = \frac{1}{\text{cos θ}} \space [\because \text{sin}^2 \text{ θ} + \text{ cos}^2 \text{ θ} = 1]\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{\text{sin}^2 \text{ θ}}{\text{cos θ}} + \text{cos θ} = \frac{1}{\text{cos θ}}.$

Let ABC be a right angled triangle with ∠C = 90°.

Given,

tan A = $\frac{3}{4}$

By formula,

$\text{tan A} = \frac{\text{Perpendicular}}{\text{Base}}$

⇒ $\frac{3}{4} = \frac{BC}{AC}$

Let BC = 3x and AC = 4x.

In △ABC,

⇒ AB² = AC² + BC²

⇒ AB² = (4x)² + (3x)²

⇒ AB² = 16x² + 9x²

⇒ AB² = 25x²

⇒ AB = $\sqrt{25x^2}$

⇒ AB = 5x.

By formula,

$$\begin{aligned}\text{sin A} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{3x}{5x} = \frac{3}{5} \\[1em] \text{sin B} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{4x}{5x} = \frac{4}{5} \\[1em] \text{cos A} = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AC}{AB} = \frac{4x}{5x} = \frac{4}{5} \\[1em] \text{cos B} = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{AB} = \frac{3x}{5x} = \frac{3}{5}.\end{aligned}$$

Substituting values in L.H.S. of sin A cos B + cos A sin B = 1.

$$\begin{aligned}\text{sin A cos B + cos A sin B} = \frac{3}{5} \times \frac{3}{5} + \frac{4}{5} \times \frac{4}{5} \\[1em] = \frac{9}{25} + \frac{16}{25} \\[1em] = \frac{9 + 16}{25} \\[1em] = \frac{25}{25} \\[1em] = 1\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that sin A cos B + cos A sin B = 1.

By formula, $\text{tan θ} = \frac{\text{Perpendicular}}{\text{Base}}$

In ΔABC,

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{BC}{AB} \\[1em] \Rightarrow \text{tan θ} = \frac{5}{12}................(1)\end{aligned}$$

In ΔADE,

$\Rightarrow \text{tan θ} = \frac{DE}{AD}$

Substituting the value of tan θ from equation (1), we get

$\Rightarrow \frac{DE}{AD} = \frac{5}{12}$.

Hence, $\frac{DE}{AD} = \frac{5}{12}$.

In right angle ∆ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 12² + 5²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$ = 13.

Let ∠CBD = x.

∠DBA = 90° - x

In ∆DAB,

⇒ ∠DAB + ∠ADB + ∠DBA = 180° [Angle sum property of triangle]

⇒ ∠DAB + 90° + 90° - x = 180°

⇒ ∠DAB = 180° - 180° + x

⇒ ∠DAB = x.

From figure,

∠DAB = ∠CAB = x

∴ ∠CBD = ∠CAB = x

(i) cos ∠CBD = cos ∠CAB

= $\frac{AB}{AC} = \frac{12}{13}$.

Hence, cos ∠CBD = $\frac{12}{13}$.

(ii) In ∆BCD,

⇒ ∠DBC + ∠DCB + ∠CDB = 180° [Angle sum property of triangle]

⇒ ∠DCB + x + 90° = 180°

⇒ ∠DCB = 180° - 90° - x

⇒ ∠DCB = 90° - x.

From figure,

∠DCB = ∠ACB =90° - x

∴ ∠ABD = ∠ACB = 90° - x

∴ cot ∠ABD = cot ∠ACB

= $\frac{BC}{AB} = \frac{5}{12}$.

Hence, cot ∠ABD = $\frac{5}{12}$.

Given,

$\text{cot α} = \frac{3}{2}$ ...........(1)

$\text{cot α} = \frac{\text{Base}}{\text{Perpendicular}} = \frac{CD}{AD}$ ........(2)

From (1) and (2) we get,

$\frac{CD}{AD} = \frac{3}{2}$

Let CD = 3x and AD = 2x.

In right angle triangle ADC,

$$\begin{aligned}\Rightarrow AC^2 = AD^2 + CD^2 \\[1em] \Rightarrow 15^2 = (2x)^2 + (3x)^2 \\[1em] \Rightarrow 225 = 4x^2 + 9x^2 \\[1em] \Rightarrow 13x^2 = 225 \\[1em] \Rightarrow x^2 = \frac{225}{13} \\[1em] \Rightarrow x = \sqrt{\frac{225}{13}} \\[1em] \Rightarrow x = \frac{15}{\sqrt{13}}.\end{aligned}$$

Perimeter of rectangle (P) = 2(CD + AD)

Area of rectangle (A) = CD × AD

Substituting values we get :

$$\begin{aligned}P = 2 \times (3x + 2x) \\[1em] = 2 \times 5x \\[1em] = 10x \\[1em] = 10 \times \frac{15}{\sqrt{13}} \text{ cm}. \\[1em] = \frac{150}{\sqrt{13}} \text{ cm}. \\[1em] A = 3x \times 2x \\[1em] = 6x^2 \\[1em] = 6 \times \Big(\frac{15}{\sqrt{13}}\Big)^2 \\[1em] = 6 \times \frac{225}{13} \\[1em] = \frac{1350}{13} \\[1em] = 103\frac{11}{13} \text{ cm}^2.\end{aligned}$$

Hence, perimeter of rectangle = $\frac{150}{\sqrt{13}}$ cm and area = $103\frac{11}{13}$ cm².

(a) In right-angled ∆BCD

Using pythagoras theorem we get :

⇒ BD² = BC² + CD²

⇒ CD² = BD² - BC²

⇒ CD² = (13)² - (12)²

⇒ CD² = 169 - 144

⇒ CD² = 25

⇒ CD = $\sqrt{25}$ = 5.

(i) By formula,

sin Φ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

In right-angled ∆BCD

sin Φ = $\frac{CD}{BD} = \frac{5}{13}$.

Hence, sin Φ = $\frac{5}{13}$.

(ii) Draw DE perpendicular to AB.

From figure,

ED = BC = 12

In right-angled ∆BED

Using pythagoras theorem we get :

⇒ BD² = ED² + EB²

⇒ 13² = 12² + EB²

⇒ EB² = (13)² - (12)²

⇒ EB² = 169 - 144

⇒ EB² = 25

⇒ EB = $\sqrt{25}$ = 5.

From figure,

AE = AB - EB = 14 - 5 = 9.

By formula,

tan θ = $\frac{\text{Perpendicular}}{\text{Base}} = \frac{ED}{AE} = \frac{12}{9} = \frac{4}{3}$.

Hence, tan θ = $\frac{4}{3}$.

(b) In right-angled ∆AED

$$\begin{aligned}\phantom{\Rightarrow} \text{sin θ} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] \Rightarrow \text{sin θ} = \frac{ED}{AD} \\[1em] \Rightarrow \text{sin θ} = \frac{12}{AD} \\[1em] \Rightarrow AD = \frac{12}{\text{sin θ}}. \\[1em] \phantom{\Rightarrow} \text{cos θ} = \frac{\text{Base}}{\text{Hypotenuse}} \\[1em] \Rightarrow \text{cos θ} = \frac{AE}{AD} \\[1em] \Rightarrow \text{cos θ} = \frac{9}{AD} \\[1em] \Rightarrow AD = \frac{9}{\text{cos θ}}.\end{aligned}$$

Hence, AD = $\frac{9}{\text{cos θ}} \text{ or } \frac{12}{\text{sin θ}}$.

Solving L.H.S. of equation : (sin A + cos A)² + (sin A - cos A)² = 2, we get,

⇒ (sin A + cos A)² + (sin A - cos A)²

⇒ sin² A + cos² A + 2 sin A cos A + sin² A + cos² A - 2 sin A cos A

Since, sin² A + cos² A = 1.

⇒ 1 + 2 sin A cos A + 1 - 2 sin A cos A

⇒ 2.

Since, L.H.S. = R.H.S.

Hence, proved that (sin A + cos A)² + (sin A - cos A)² = 2.

Solving L.H.S. of the equation : cot² A - $\frac{1}{\text{sin}^2 A}$ + 1 = 0

$$\begin{aligned}\Rightarrow \frac{\text{cos}^2 A}{\text{sin}^2 A} - \frac{1}{\text{sin}^2 A} + 1 \\[1em] \Rightarrow \frac{\text{cos}^2 A - 1 + \text{sin}^2 A}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{\text{sin}^2 A + \text{cos}^2 A - 1}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{1 - 1}{\text{sin}^2 A} \\[1em] \Rightarrow \frac{0}{\text{sin}^2 A} \\[1em] \Rightarrow 0.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\text{cot}^2 A - \frac{1}{\text{sin}^2 A} + 1 = 0$

Solving L.H.S. of the equation : $\frac{1}{\text{1 + tan}^2 A} + \frac{1}{\text{1 + cot}^2 A}$ = 1.

$$\begin{aligned}\phantom{=} \frac{1}{1 + \frac{\text{sin}^2 A}{\text{cos}^2 A}} + \frac{1}{1 + \frac{\text{cos}^2 A}{\text{sin}^2 A}} \\[1em] = \frac{1}{\frac{\text{cos}^2 A + \text{sin}^2 A}{\text{cos}^2 A}} + \frac{1}{\frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A}} \\[1em] = \frac{\text{cos}^2 A}{\text{sin}^2 A + \text{cos}^2 A} + \frac{\text{sin}^2 A}{\text{sin}^2 A + \text{cos}^2 A} \\[1em] = \frac{\text{sin}^2 A + \text{cos}^2 A}{\text{sin}^2 A + \text{cos}^2 A} \\[1em] = 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that $\frac{1}{\text{1 + tan}^2 A} + \frac{1}{\text{1 + cot}^2 A}$ = 1.

Substituting, 1 - sin² θ = cos² θ and 1 - cos² θ = sin² θ in $\sqrt{\frac{1 - \text{sin}^2 \text{ θ}}{1 - \text{cos}^2 \text{ θ}}}$ we get :

$$\begin{aligned}\Rightarrow \sqrt{\frac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}} \\[1em] \Rightarrow \sqrt{\text{cot}^2 \text{ θ}} \\[1em] \Rightarrow \text{cot θ}.\end{aligned}$$

Hence, $\sqrt{\frac{1 - \text{sin}^2 \text{ θ}}{1 - \text{cos}^2 \text{ θ}}}$ = cot θ.

Given,

⇒ sin θ + cosec θ = 2

Squaring both sides we get :

⇒ (sin θ + cosec θ)² = 2²

⇒ sin² θ + cosec² θ + 2 sin θ. cosec θ = 4

⇒ sin² θ + cosec² θ + $2\text{ sin θ} \times \frac{1}{\text{sin θ}} = 4$

⇒ sin² θ + cosec² θ + 2 = 4

⇒ sin² θ + cosec² θ = 4 - 2

⇒ sin² θ + cosec² θ = 2.

Hence, proved that sin² θ + cosec² θ = 2.

x² + y² = (a cos θ + b sin θ)² + (a sin θ - b cos θ)²

⇒ x² + y² = a² cos² θ + b² sin² θ + 2ab cos θ. sin θ + a² sin² θ + b² cos² θ - 2ab cos θ. sin θ

⇒ x² + y² = a²(sin² θ + cos² θ) + b²(sin² θ + cos² θ)

⇒ x² + y² = (a² + b²)(sin² θ + cos² θ)

As, sin² θ + cos² θ = 1.

⇒ x² + y² = (a² + b²).

Hence proved that x² + y² = (a² + b²).

In right angled triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (24)² + (7)²

⇒ AC² = 576 + 49

⇒ AC² = 625

⇒ AC = $\sqrt{625}$ = 25 cm.

By formula,

sin A = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{BC}{AC} = \frac{7}{25}$.

Hence, Option 2 is the correct option.

By formula,

sec A = $\frac{\text{Hypotenuse}}{\text{Base}}$

= $\frac{AC}{AB} = \frac{25}{24}$.

Hence, Option 3 is the correct option.

By formula,

tan C = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{AB}{BC} = \frac{24}{7}$.

Hence, Option 1 is the correct option.

By formula,

cosec C = $\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

= $\frac{AC}{AB} = \frac{25}{24}$.

Hence, Option 4 is the correct option.

By formula,

tan A = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{BC}{AB} = \frac{7}{24}$.

cot C = $\frac{\text{Base}}{\text{Perpendicular}}$

= $\frac{BC}{AB} = \frac{7}{24}$.

Substituting value in tan A + cot C we get :

$$\begin{aligned}\text{tan A + cot C} = \frac{7}{24} + \frac{7}{24} \\[1em] = \frac{14}{24} \\[1em] = \frac{7}{12}.\end{aligned}$$

Hence, Option 1 is the correct option.

By formula,

cos A = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AB}{AC} = \frac{24}{25}$.

sin C = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{AB}{AC} = \frac{24}{25}$.

Substituting value in 2 cos A - sin C we get :

$$\begin{aligned}\text{2 cos A - sin C} = 2 \times \frac{24}{25} - \frac{24}{25} \\[1em] = \frac{48}{25} - \frac{24}{25} \\[1em] = \frac{24}{25}.\end{aligned}$$

Hence, Option 2 is the correct option.

In right angle triangle ABC,

⇒ BC² = AB² + AC²

⇒ 10² = (6)² + (AC)²

⇒ AC² = 10² - 6²

⇒ AC² = 100 - 36

⇒ AC² = 64

⇒ AC = $\sqrt{64}$ = 8 cm.

By formula,

sin B = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{AC}{BC} = \frac{8}{10}$.

sin C = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

= $\frac{AB}{BC} = \frac{6}{10}$.

cos B = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AB}{BC} = \frac{6}{10}$.

cos C = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AC}{BC} = \frac{8}{10}$.

Substituting values in sin B cos C + sin C cos B we get :

$$\begin{aligned}\Rightarrow \text{sin B cos C + sin C cos B} = \frac{8}{10} \times \frac{8}{10} + \frac{6}{10} \times \frac{6}{10} \\[1em] = \frac{64}{100} + \frac{36}{100} \\[1em] = \frac{100}{100} = 1.\end{aligned}$$

Hence, Option 2 is the correct option.

In right angled triangle BDC,

⇒ BC² = BD² + CD²

⇒ BC² = (4)² + (3)²

⇒ BC² = 16 + 9

⇒ BC² = 25

⇒ BC = $\sqrt{25}$ = 5 cm.

In right angled triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (12)² + (5)²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$ = 13 cm.

By formula,

cos θ = $\frac{\text{Base}}{\text{Hypotenuse}}$

= $\frac{AB}{AC} = \frac{12}{13}$.

Hence, Option 1 is the correct option.

Let ABC be a right angle triangle with ∠B = 90°.

By formula,

cos A = $\frac{\text{Base}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \frac{4}{5} = \frac{AB}{AC}$

Let AB = 4x and AC = 5x.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (5x)² = (4x)² + BC²

⇒ 25x² = 16x² + BC²

⇒ BC² = 25x² - 16x²

⇒ BC² = 9x²

⇒ BC = $\sqrt{9x^2}$ = 3x.

By formula,

tan A = $\frac{\text{Perpendicular}}{\text{Base}}$

= $\frac{BC}{AB} = \frac{3x}{4x} = \frac{3}{4}$.

Hence, Option 2 is the correct option.

Let ABC be a right angle triangle with ∠B = 90°.

By formula,

sin A = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \frac{1}{2} = \frac{BC}{AC}$

Let BC = x and AC = 2x.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (2x)² = AB² + (x)²

⇒ 4x² = AB² + x²

⇒ AB² = 3x²

⇒ AB = $\sqrt{3x^2} = \sqrt{3}x$.

By formula,

cot A = $\frac{\text{Base}}{\text{Perpendicular}} = \frac{AB}{BC} = \frac{\sqrt{3}x}{x} = \sqrt{3}$.

Hence, Option 1 is the correct option.

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

By formula,

cosec θ = $\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

Substituting values we get :

$\Rightarrow \frac{13}{12} = \frac{AC}{AB}$

Let AC = 13x and AB = 12x.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (13x)² = (12x)² + BC²

⇒ 169x² = 144x² + BC²

⇒ BC² = 169x² - 144x²

⇒ BC² = 25x²

⇒ BC = $\sqrt{25x^2} = 5x$.

By formula,

tan θ = $\frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC} = \frac{12x}{5x} = \frac{12}{5}$.

Hence, Option 1 is the correct option.

Let ABC be a right angle triangle with ∠B = 90°.

By formula,

tan A = $\frac{\text{Perpendicular}}{\text{Base}}$

Substituting values we get :

$\Rightarrow \frac{x}{y} = \frac{BC}{AB}$

Let BC = xk and AB = yk.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (yk)² + (xk)²

⇒ AC² = y²k² + x²k²

⇒ AC² = k²(y² + x²)

⇒ AC = $\sqrt{k^2(y^2 + x^2)}$

⇒ AC = $k\sqrt{y^2 + x^2}$.

By formula,

cos A = $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{yk}{k\sqrt{x^2 + y^2}} = \frac{y}{\sqrt{x^2 + y^2}}$.

Hence, Option 2 is the correct option.

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

By formula,

sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \frac{a}{b} = \frac{AB}{AC}$

Let AB = ak and AC = bk.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (bk)² = (ak)² + BC²

⇒ b²k² = a²k² + BC²

⇒ BC² = b²k² - a²k²

⇒ BC² = $\sqrt{k^2(b^2 - a^2)}$

⇒ BC = $k\sqrt{(b^2 - a^2)}$.

By formula,

$$\begin{aligned}\text{cos θ} = \frac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \frac{BC}{AC} \\[1em] = \frac{k\sqrt{(b^2 - a^2)}}{bk} \\[1em] = \frac{\sqrt{b^2 - a^2}}{b}\end{aligned}$$

Hence, Option 3 is the correct option.

Given, sin A = $\frac{1}{2}$

By formula,

⇒ sin² A + cos² A = 1

⇒ $\Big(\frac{1}{2}\Big)^2$ + cos² A = 1

⇒ $\frac{1}{4}$ + cos² A = 1

⇒ cos² A = $1 - \frac{1}{4}$

⇒ cos² A = $\frac{3}{4}$

⇒ cos A = $\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

By formula,

⇒ cot A = $\frac{\text{cos A}}{\text{sin A}} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.

Thus, both the statements are false.

Hence, option 2 is the correct option.

Given, tan A = $\frac{1}{\sqrt{3}}$

As we know,

$$\begin{aligned}\text{tan A }= \frac{\text{Perpendicular}}{\text{Base}}\\[1em] \Rightarrow \frac{BC}{AB} = \frac{1}{\sqrt{3}}\end{aligned}$$

Let BC = k and AB = $\sqrt{3}$ k

Since, ΔABC is a right angled triangle, using pythagoras theorem,

⇒ AC² = AB² + BC²

∴ Reason (R) is true.

Substituting values we get :

$$\begin{aligned}\Rightarrow AC^2 = (\sqrt{3}k)^2 + k^2\\[1em] \Rightarrow AC^2 = 3k^2 + k^2\\[1em] \Rightarrow AC^2 = 4k^2\\[1em] \Rightarrow AC = \sqrt{4k^2}\\[1em] \Rightarrow AC = 2k \\[1em] \Rightarrow AC = 2k\frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AC = 2k \times \frac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow AC = \frac{2(\sqrt{3}k)}{\sqrt{3}} \\[1em] \Rightarrow AC = \frac{2AB}{\sqrt{3}}.\end{aligned}$$

∴ Assertion (A) is false.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

In triangle ABC,

Using pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 4² + 3²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

By formula,

$$\begin{aligned}\text{sin A} = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\\[1em] = \frac{BC}{AC}\\[1em] = \frac{3}{5}.\end{aligned}$$

By formula,

$$\begin{aligned}\text{cos A} = \frac{\text{Base}}{\text{Hypotenuse}}\\[1em] = \frac{AB}{AC}\\[1em] = \frac{4}{5}.\end{aligned}$$

Substituting values we get,

sin A cos A = $\frac{3}{5} \times \frac{4}{5} = \frac{3 \times 4}{5 \times 5} = \frac{12}{25}$.

∴ Assertion (A) is true.

By formula,

$\frac{1}{\text{sin A}}$ = cosec A ≠ cos A

∴ Reason (R) is false.

∴ Assertion (A) is true, Reason (R) is false.

Hence, option 1 is the correct option.

Given, x = a cos θ + b sin θ and y = a cos θ - b sin θ

⇒ x² = (a cos θ + b sin θ)²

⇒ x² = a² cos² θ + b² sin² θ + 2 ab cos θ sin θ .....................(1)

Similarly,

⇒ y² = (a cos θ - b sin θ)²

⇒ y² = a² cos² θ + b² sin² θ - 2 a b cos θ sin θ .....................(2)

Adding equations (1) and (2), we get :

⇒ x² + y² = (a² cos² θ + b² sin² θ + 2ab cos θ sin θ) + (a² cos² θ + b² sin² θ - 2ab cos θ sin θ)

⇒ x² + y² = (a² cos² θ + a² cos² θ) + (b² sin² θ + b² sin² θ) + (2ab cos θ sin θ - 2ab cos θ sin θ)

⇒ x² + y² = 2a² cos² θ + 2 b² sin² θ

⇒ x² + y² = 2(a² cos² θ + b² sin² θ)

∴ Assertion (A) is false.

cos² θ + sin² θ = 1

This is a fundamental Pythagorean trigonometric identity and is always true for any real value of θ.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

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,

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

For angle C,

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

From figure,

$$\begin{aligned}\Rightarrow \text{cot θ} = \frac{\text{Base}}{\text{Perpendicular}} \\[1em] \Rightarrow \text{cot θ} = \frac{AB}{BC} \\[1em] \Rightarrow \text{cot θ} = \frac{x}{10} \\[1em] \Rightarrow x = 10 \text{ cot θ}. \\[1em] \Rightarrow \text{cosec θ} = \frac{\text{Hypotenuse}}{\text{Perpendicular}} \\[1em] \Rightarrow \text{cosec θ} = \frac{AC}{BC} \\[1em] \Rightarrow \text{cosec θ} = \frac{y}{10} \\[1em] \Rightarrow y = 10 \text{ cosec θ}.\end{aligned}$$

Hence, x = 10 cot θ and y = 10 cosec θ.

(i) In right angle triangle ABC,

By pythagoras theorem we get :

⇒ AB² = AC² + BC²

⇒ 20² = AC² + 12²

⇒ 400 = AC² + 144

⇒ AC² = 400 - 144

⇒ AC² = 256

⇒ AC = $\sqrt{256}$ = 16.

$$\begin{aligned}\text{sin ∠ABC} = \frac{\text{Perpendicular}}{\text{\text{Hypotenuse}}} \\[1em] = \frac{AC}{AB} \\[1em] = \frac{16}{20} = \frac{4}{5}.\end{aligned}$$

Hence, sin ∠ABC = $\frac{4}{5}$.

(ii) In right angle triangle BCD,

By pythagoras theorem we get :

⇒ BD² = BC² + CD²

⇒ BD² = 12² + 9²

⇒ BD² = 144 + 81

⇒ BD² = 225

⇒ BD = $\sqrt{225}$

⇒ BD = 15.

By formula,

$$\begin{aligned}\text{tan x } = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] = \frac{BC}{CD} = \frac{12}{9} = \frac{4}{3}. \\[1em] \text{cos x } = \frac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \frac{CD}{BD} = \frac{9}{15} = \frac{3}{5}. \\[1em] \text{sin x } = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \frac{BC}{BD} = \frac{12}{15} = \frac{4}{5}. \\[1em]\end{aligned}$$

Substituting values in tan x - cos x + 3 sin x we get :

$$\begin{aligned}\Rightarrow \frac{4}{3} - \frac{3}{5} + 3 \times \frac{4}{5} \\[1em] \Rightarrow \frac{4}{3} - \frac{3}{5} + \frac{12}{5} \\[1em] \Rightarrow \frac{20 - 9 + 36}{15} \\[1em] \Rightarrow \frac{47}{15} \\[1em] \Rightarrow 3\frac{2}{15}.\end{aligned}$$

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

(i) By formula,

$$\begin{aligned}\text{sin x } = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\[1em] = \frac{AD}{AB} = \frac{15}{25} = \frac{3}{5}. \\[1em] 5\text{ sin x} = 5 \times \frac{3}{5} = 3.\end{aligned}$$

Hence, 5 sin x = 3.

(ii) In right angle triangle ABD,

⇒ AB² = AD² + BD²

⇒ 25² = 15² + BD²

⇒ 625 = 225 + BD²

⇒ BD² = 625 - 225

⇒ BD² = 400

⇒ BD = $\sqrt{400}$ = 20.

By formula,

$$\begin{aligned}\text{tan x } = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] = \frac{AD}{BD} = \frac{15}{20} = \frac{3}{4}. \\[1em] 7\text{ tan x} = 7 \times \frac{3}{4} = \frac{21}{4} = 5\frac{1}{4}.\end{aligned}$$

Hence, 7 tan x = $5\frac{1}{4}$.

(iii) In right angle triangle ADC,

⇒ AC² = AD² + DC²

⇒ 17² = 15² + DC²

⇒ 289 = 225 + DC²

⇒ DC² = 289 - 225

⇒ DC² = 64

⇒ DC = $\sqrt{64}$ = 8.

Solving,

$$\begin{aligned}\text{5 cos x - 17 sin y - tan x} = 5 \times \frac{BD}{AB} - 17 \times \frac{CD}{AC} - \frac{AD}{BD} \\[1em] = 5 \times \frac{20}{25} - 17 \times \frac{8}{17} - \frac{15}{20} \\[1em] = 4 - 8 - \frac{3}{4} \\[1em] = -4 - \frac{3}{4} \\[1em] = \frac{-16 - 3}{4}\\[1em] = -\frac{19}{4} \\[1em] = -4\frac{3}{4}.\end{aligned}$$

Hence, 5 cos x - 17 sin y - tan x = -$4\frac{3}{4}.$

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

Given,

⇒ q cos θ = p

⇒ cos θ = $\frac{p}{q}$ ..........(1)

By formula,

⇒ cos θ = $\frac{\text{Base}}{\text{Hypotenuse}}$

⇒ cos θ = $\frac{BC}{AC}$ ..........(2)

Comparing equations (1) and (2) we get :

$\Rightarrow \frac{p}{q} = \frac{BC}{AC}$

Let BC = pk and AC = qk.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (qk)² = AB² + (pk)²

⇒ AB² = q²k² - p²k²

⇒ AB² = k²(q² - p²)

⇒ AB = $\sqrt{k^2(q^2 - p^2)} = k\sqrt{q^2 - p^2}$.

By formula,

$$\begin{aligned}\text{tan θ} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] = \frac{AB}{BC} = \frac{k\sqrt{q^2 - p^2}}{pk} \\[1em] = \frac{\sqrt{q^2 - p^2}}{p}. \\[1em] \text{cot θ} = \frac{1}{\text{tan θ}} \\[1em] = \frac{1}{\frac{\sqrt{q^2 - p^2}}{p}} \\[1em] = \frac{p}{\sqrt{q^2 - p^2}}.\end{aligned}$$

Substituting values in tan θ - cot θ we get :

$$\begin{aligned}\Rightarrow \text{tan θ - cot θ} = \frac{\sqrt{q^2 - p^2}}{p} - \frac{p}{\sqrt{q^2 - p^2}} \\[1em] = \frac{\sqrt{q^2 - p^2}\sqrt{q^2 - p^2} - p^2}{p\sqrt{q^2 - p^2}} \\[1em] = \frac{q^2 - p^2 - p^2}{p\sqrt{q^2 - p^2}} \\[1em] = \frac{q^2 - 2p^2}{p\sqrt{q^2 - p^2}}.\end{aligned}$$

Hence, tan θ - cot θ = $\frac{q^2 - 2p^2}{p\sqrt{q^2 - p^2}}.$

Let ABC be a triangle with ∠B = 90° and ∠C = θ.

Given,

⇒ 4 sin θ = 3 cos θ

⇒ $\frac{\text{sin θ}}{\text{cos θ}} = \frac{3}{4}$.

⇒ tan θ = $\frac{3}{4}$ ...........(1)

From figure.

⇒ tan θ = $\frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC}$ ............(2)

From (1) and (2) we get :

$\frac{AB}{BC} = \frac{3}{4}$

Let AB = 3x and BC = 4x.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC² = $\sqrt{25x^2}$

⇒ AC = 5x.

(i) By formula,

⇒ sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{3x}{5x} = \frac{3}{5}$.

Hence, sin θ = $\frac{3}{5}$.

(ii) By formula,

⇒ cos θ = $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{4x}{5x} = \frac{4}{5}$.

Hence, cos θ = $\frac{4}{5}$.

(iii) By formula,

⇒ cot θ = $\frac{\text{Base}}{\text{Perpendicular}} = \frac{BC}{AB} = \frac{4x}{3x} = \frac{4}{3}$.

⇒ cot² θ = $\Big(\frac{4}{3}\Big)^2$ = $\frac{16}{9}$.

⇒ cosec θ = $\frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{AC}{AB} = \frac{5x}{3x} = \frac{5}{3}$.

⇒ cosec² θ = $\Big(\frac{5}{3}\Big)^2$ = $\frac{25}{9}$.

$$\begin{aligned}\text{cot}^2 \text{ θ} - \text{cosec}^2 \text{ θ} = \frac{16}{9} - \frac{25}{9} \\[1em] = \frac{16 - 25}{9} \\[1em] = -\frac{9}{9} \\[1em] = -1.\end{aligned}$$

Hence, cot² θ - cosec² θ = -1.

Given,

⇒ 2 cos θ = $\sqrt{3}$

⇒ cos θ = $\frac{\sqrt{3}}{2}$

Squaring both sides we get :

⇒ cos² θ = $\Big(\frac{\sqrt{3}}{2}\Big)^2$ = $\frac{3}{4}$.

⇒ 1 - sin² θ = $\frac{3}{4}$.

⇒ sin² θ = $1 - \frac{3}{4}$.

⇒ sin² θ = $\frac{4 - 3}{4}$.

⇒ sin² θ = $\frac{1}{4}$.

⇒ sin θ = $\sqrt{\frac{1}{4}}$

⇒ sin θ = $\frac{1}{2}$.

Substituting values in L.H.S. of equation 3 sin θ - 4 sin³ θ = 1 we get :

$$\begin{aligned}\Rightarrow 3\text{sin θ} - 4\text{ sin}^3 θ \\[1em] \Rightarrow \text{sin θ }(3 - 4\text{ sin}^2 θ) \\[1em] \Rightarrow \frac{1}{2} \times (3 - 4 \times \frac{1}{4}) \\[1em] \Rightarrow \frac{1}{2} \times (3 - 1) \\[1em] \Rightarrow \frac{1}{2} \times 2 \\[1em] \Rightarrow 1.\end{aligned}$$

Since, L.H.S. = R.H.S.

Hence, proved that 3sin θ - 4 sin³ θ = 1.

Given, $\frac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} = \frac{1}{4}.$

Solving above equation we get,

$$\begin{aligned}\Rightarrow \frac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} = \frac{1}{4} \\[1em] \Rightarrow \frac{\frac{1}{\text{cos θ}} - \frac{\text{sin θ}}{\text{cos θ}}}{\frac{1}{\text{cos θ}} + \frac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \frac{\frac{\text{1 - sin θ}}{\text{cos θ}}}{\frac{\text{1 + sin θ}}{{\text{cos θ}}}} \\[1em] \Rightarrow \frac{\text{1 - sin θ}}{\text{1 + sin θ}} = \frac{1}{4} \\[1em] \Rightarrow 4(1 - \text{sin θ}) = 1 + \text{sin θ} \\[1em] \Rightarrow 4 - 4\text{ sin θ} = 1 + \text{sin θ} \\[1em] \Rightarrow \text{sin θ + 4 sin θ} = 4 - 1 \\[1em] \Rightarrow 5 \text{ sin θ} = 3 \\[1em] \Rightarrow \text{sin θ} = \frac{3}{5}.\end{aligned}$$

Hence, sin θ = $\frac{3}{5}$.

Given,

$$\begin{aligned}\phantom{\Rightarrow} \text{sin θ + cosec θ} = 3\frac{1}{3} \\[1em] \Rightarrow \text{sin θ + cosec θ} = \frac{10}{3} \\[1em]\end{aligned}$$

Squaring both sides we get,

$$\begin{aligned}\Rightarrow \text{(sin θ + cosec θ)}^2 = \Big(\frac{10}{3}\Big)^2 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{ 2 sin θ.cosec θ} = \frac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + \text{2 sin θ} \times \frac{1}{\text{sin θ}} = \frac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} + 2 = \frac{100}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \frac{100}{9} - 2 \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \frac{100 - 18}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = \frac{82}{9} \\[1em] \Rightarrow \text{sin}^2 \text{ θ} + \text{cosec}^2 \text{ θ} = 9\frac{1}{9}\end{aligned}$$

Hence, sin² θ + cosec² θ = $9\frac{1}{9}$.

By formula,

sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

cosec θ = $\frac{\text{Hypotenuse}}{\text{Perpendicular}}$

In ΔABD,

$$\begin{aligned}\Rightarrow \text{cosec x} = \frac{AB}{BD}\\[1em] \Rightarrow \frac{13}{5} = \frac{26}{BD}\\[1em] \Rightarrow BD = \frac{26 \times 5}{13}\\[1em] \Rightarrow BD = 2 \times 5\\[1em] \Rightarrow BD = 10 \text{ cm}.\end{aligned}$$

Since, ΔABD is a right angled triangle. Using pythagoras theorem,

⇒ AB² = BD² + AD²

⇒ 26² = 10² + AD²

⇒ 676 = 100 + AD²

⇒ AD² = 676 - 100

⇒ AD² = 576

⇒ AD = $\sqrt{576}$

⇒ AD = ± 24 cm

As length of side of a triangle cannot be negative. So, AD = 24 cm.

In ΔADC,

$$\begin{aligned}\Rightarrow \text{sin y} = \frac{AD}{AC}\\[1em] \Rightarrow \frac{8}{17} = \frac{24}{AC}\\[1em] \Rightarrow AC = \frac{24 \times 17}{8}\\[1em] \Rightarrow AC = 3 \times 17\\[1em] \Rightarrow AC = 51 \text{ cm}.\end{aligned}$$

Since, ΔADC is a right angled triangle. Using pythagoras theorem,

⇒ AC² = AD² + DC²

⇒ 51² = 24² + DC²

⇒ 2601 = 576 + DC²

⇒ DC² = 2601 - 576

⇒ DC² = 2025

⇒ DC = $\sqrt{2025}$

⇒ DC = ± 45

As length of side of a triangle cannot be negative. So, DC = 45 cm.

From figure,

BC = BD + DC = 10 + 45 = 55 cm.

Hence, the length of BC = 55 cm.

In △ABC,

By formula,

sin α = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{3}{5} = \frac{AB}{AC} \\[1em] \Rightarrow \frac{3}{5} = \frac{4}{AC} \\[1em] \Rightarrow AC = \frac{20}{3}\end{aligned}$$

In right angle triangle ABC,

By pythagoras theorem, we get :

⇒ AC² = AB² + BC²

$$\begin{aligned}\Rightarrow \Big(\frac{20}{3}\Big)^2 = 4^2 + BC^2 \\[1em] \Rightarrow \frac{400}{9} = 16 + BC^2 \\[1em] \Rightarrow BC^2 = \frac{400}{9} - 16 \\[1em] \Rightarrow BC^2 = \frac{400 - 144}{9} \\[1em] \Rightarrow BC^2 = \frac{256}{9} \\[1em] \Rightarrow BC = \sqrt{\frac{256}{16}} \\[1em] \Rightarrow BC = \frac{16}{3} \text{ m}\end{aligned}$$

In △CDE,

By formula,

cos β = $\frac{\text{Base}}{\text{Hypotenuse}}$

Substituting values we get :

$\Rightarrow \frac{12}{13} = \frac{CD}{CE}$

Let CD = 12k and CE = 13k.

In right angle triangle ABC,

⇒ CE² = CD² + ED²

⇒ (13k)² = (12k)² + 3²

⇒ 169k² = 144k² + 3²

⇒ 3² = 169k² - 144k²

⇒ 9 = $25k^2$

⇒ $k = \sqrt{\frac{9}{25}} = \frac{3}{5}$.

CD = 12k = $12 \times \frac{3}{5} = \frac{36}{5}$

From figure,

$$\begin{aligned}BD = BC + CD = \frac{16}{3} + \frac{36}{5} \\[1em] = \frac{80 + 108}{15} \\[1em] = \frac{188}{15} \\[1em] = 12\frac{8}{15} \text{ m}.\end{aligned}$$

Hence, BD = $12\frac{8}{15}$ m.