Trigonometrical Ratios — Exercise 21(A)

If sin $A = \dfrac{5}{13}$ the value of tan A is :

1. $\dfrac{5}{12}$

2. $\dfrac{12}{13}$

3. $\dfrac{12}{5}$

4. $\dfrac{13}{12}$

Answer

Given:

sin $A = \dfrac{5}{13}$

i.e., $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{5}{13}$

∴ If length of BC = 5x unit, length of AC = 13x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (13x)² = AB² + (5x)²

⇒ 169x² = AB² + 25x²

⇒ AB² = 169x² - 25x²

⇒ AB² = 144x²

⇒ AB = $\sqrt{144x^2}$

⇒ AB = 12x

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

= $\dfrac{BC}{AB}$ = $\dfrac{5x}{12x}$ = $\dfrac{5}{12}$

Hence, option 1 is the correct option.

If tan $A = \dfrac{3}{5}$, the value of sin² A + cos² A is :

1. $\dfrac{9}{25}$

2. 1

3. $\dfrac{9}{16}$

4. $\dfrac{16}{9}$

Answer

Given:

tan $A = \dfrac{3}{5}$

i.e. $\dfrac{Perpendicular}{Base} = \dfrac{3}{5}$

∴ If length of BC = 3x unit, length of AB = 5x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (3x)²

⇒ AC² = 25x² + 9x²

⇒ AC² = 34x²

⇒ AC = $\sqrt{34\text{x}^2}$

⇒ AC = $\sqrt{34}$ x

sin $A = \dfrac{Perpendicular}{Hypotenuse}$ =

$= \dfrac{BC}{AC} = \dfrac{3x}{\sqrt{34}x} = \dfrac{3}{\sqrt{34}}$

cos $A = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC} = \dfrac{5x}{\sqrt{34}x} = \dfrac{5}{\sqrt{34}}$

Now, sin² A + cos² A

$= \Big(\dfrac{3}{\sqrt{34}}\Big)^2 + \Big(\dfrac{5}{\sqrt{34}}\Big)^2\\[1em] = \Big(\dfrac{9}{34}\Big) + \Big(\dfrac{25}{34}\Big)\\[1em] = \Big(\dfrac{9 + 25}{34}\Big)\\[1em] = \Big(\dfrac{34}{34}\Big)\\[1em] = 1$

Hence, option 2 is the correct option.

If cot $A = \dfrac{5}{12}$, the value of cot² A - cosec² A is :

1. 1

2. 2

3. -2

4. -1

Answer

Given:

cot $A = \dfrac{5}{12}$

i.e. $\dfrac{Base}{Perpendicular} = \dfrac{5}{12}$

∴ If length of AB = 5x unit, length of BC = 12x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169²

⇒ AC = $\sqrt{169\text{x}^2}$

⇒ AC = 13x

cosec $A = \dfrac{Hypotenuse}{Perpendicular}$

$\dfrac{AC}{BC} = \dfrac{13x}{12x} = \dfrac{13}{12}$

Now, cot² A - cosec² A

$= \Big(\dfrac{5}{12}\Big)^2 - \Big(\dfrac{13}{12}\Big)^2\\[1em] = \Big(\dfrac{25}{144}\Big) - \Big(\dfrac{169}{144}\Big)\\[1em] = \Big(\dfrac{25 - 169}{144}\Big)\\[1em] = \Big(\dfrac{- 144}{144}\Big)\\[1em] = - 1$

Hence, option 4 is the correct option.

In the given figure (each observation is in cm) tan C is :

1. $\dfrac{3}{5}$

2. $\dfrac{4}{3}$

3. $\dfrac{3}{4}$

4. $\dfrac{4}{5}$

Answer

In Δ ABD,

AB² = AD² + BD²

⇒ (26)² = AD² + (10)²

⇒ 676 = AD² + 100

⇒ AD² = 676 - 100

⇒ AD² = 576

⇒ AD = $\sqrt{576}$

⇒ AD = 24 cm

In Δ ADC,

AC² = AD² + DC²

⇒ AC² = (24)² + (32)²

⇒ AC² = 576 + 1,024

⇒ AC² = 1,600

⇒ AC = $\sqrt{1,600}$

⇒ AC = 40 cm

tan $A = \dfrac{Perpendicular}{Base}$

= $\dfrac{24}{32}$

= $\dfrac{3}{4}$

Hence, option 3 is the correct option.

If 5 cos A = 3, the value of sec² A - tan² A is :

1. 1

2. -1

3. $\dfrac{3}{4}$

4. $\dfrac{4}{3}$

Answer

Given:

5 cos A = 3

⇒ cos A = $\dfrac{3}{5}$

i.e., $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{3}{5}$

∴ If length of AB = 3x unit, length of AC = 5x unit.

In Δ ABC,

AC² = BC² + AB²

⇒ (5x)² = BC² + (3x)²

⇒ 25x² = BC² + 9x²

⇒ BC² = 25x² - 9x²

⇒ BC² = 16x²

⇒ BC = $\sqrt{16\text{x}^2}$

⇒ BC = 4x

sec A = $\dfrac{Hypotenuse}{Base}$

= $\dfrac{AC}{BA} = \dfrac{5x}{3x} = \dfrac{5}{3}$

tan A = $\dfrac{Perpendicular}{Base}$

= $\dfrac{BC}{BA} = \dfrac{4x}{3x} = \dfrac{4}{3}$

Now, sec² A - tan² A

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

Hence, option 1 is the correct option.

From the following figure, find the values of :

(i) sin A

(ii) cos A

(iii) cot A

(iv) sec C

(v) cosec C

(vi) tan C.

Answer

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = 3² + 4²

⇒ AC² = 9 + 16

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

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

$= \dfrac{CB}{AC}\\[1em] = \dfrac{4}{5}\\[1em]$

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

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

$= \dfrac{AB}{AC}\\[1em] = \dfrac{3}{5}\\[1em]$

Hence, cos $A = \dfrac{3}{5}$.

(iii) cot $A = \dfrac{Base}{Perpendicular}$

$= \dfrac{AB}{BC}\\[1em] = \dfrac{3}{4}\\[1em]$

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

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

$= \dfrac{AC}{AB}\\[1em] = \dfrac{5}{4}\\[1em] = 1\dfrac{1}{4}\\[1em]$

Hence, sec $A = \dfrac{5}{4} = 1\dfrac{1}{4}$.

(v) cosec $C = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AC}{AB}\\[1em] = \dfrac{5}{3}\\[1em] = 1\dfrac{2}{3}\\[1em]$

Hence, cosec $C = \dfrac{5}{3} = 1\dfrac{2}{3}$.

(vi) tan $C = \dfrac{Perpendicular}{Base}$

$= \dfrac{AB}{BC}\\[1em] = \dfrac{3}{4}\\[1em]$

Hence, tan $C = \dfrac{3}{4}$.

From the following figure, find the values of :

(i) cos B

(ii) tan C

(iii) sin²B + cos²B

(iv) sin B.cos C + cos B.sin C

Answer

In Δ BAC,

⇒ BC² = AB² + AC² (∵ BC is hypotenuse)

⇒ (17)² = (8)² + AC²

⇒ 289 = 64 + AC²

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = $\sqrt{225}$

⇒ AC = 15

(i) cos $B = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{BC}\\[1em] = \dfrac{8}{17}$

Hence, cos $B = \dfrac{8}{17}$.

(ii) tan $C = \dfrac{Perpendicular}{Base}$

$= \dfrac{AB}{AC}\\[1em] = \dfrac{8}{15}\\[1em]$

Hence, tan $C = \dfrac{8}{15}$.

(iii) sin²B + cos²B

$= \Big(\dfrac{Perpendicular}{Hypotenuse}\Big)^2 + \Big(\dfrac{Base}{Hypotenuse}\Big)^2\\[1em] = \Big(\dfrac{AC}{BC}\Big)^2 + \Big(\dfrac{AB}{BC}\Big)^2\\[1em] = \Big(\dfrac{15}{17}\Big)^2 + \Big(\dfrac{8}{17}\Big)^2\\[1em] = \dfrac{225}{289} + \dfrac{64}{289}\\[1em] = \dfrac{225 + 64}{289}\\[1em] = \dfrac{289}{289}\\[1em] = 1$

Hence, sin²B + cos²B = 1.

(iv) sin B.cos C + cos B.sin C

$= \dfrac{Perpendicular}{Hypotenuse} .\dfrac{Base}{Hypotenuse} + \dfrac{Base}{Hypotenuse} . \dfrac{Perpendicular}{Hypotenuse}\\[1em] = \dfrac{AC}{BC} .\dfrac{AC}{BC} + \dfrac{AB}{BC} . \dfrac{AB}{BC}\\[1em] = \Big(\dfrac{AC}{BC}\Big)^2 + \Big(\dfrac{AB}{BC}\Big)^2\\[1em] = \Big(\dfrac{15}{17}\Big)^2 + \Big(\dfrac{8}{17}\Big)^2\\[1em] = \dfrac{225}{289} + \dfrac{64}{289}\\[1em] = \dfrac{225 + 64}{289}\\[1em] = \dfrac{289}{289}\\[1em] = 1$

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

From the following figure, find the values of :

(i) cos A

(ii) cosec A

(iii) tan²A – sec²A

(iv) sin C

(v) sec C

(v) cot²C - $\dfrac{1}{\text{sin}^2 \text{ C}}$

Answer

In Δ ADB,

⇒ AB² = AD² + BD² (∵ AB is hypotenuse)

⇒ AB² = (3)² + (4)²

⇒ AB² = 9 + 16

⇒ AB² = 25

⇒ AB = $\sqrt{25}$

⇒ AB = 5

In Δ CDB,

⇒ BC² = BD² + DC² (∵ BC is hypotenuse)

⇒ (12)² = (4)² + DC²

⇒ 144 = 16 + DC²

⇒ DC² = 144 - 16

⇒ DC² = 128

⇒ DC = $\sqrt{128}$

⇒ DC = $8\sqrt{2}$

(i) cos $A = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AD}{AB}\\[1em] = \dfrac{3}{5}$

Hence, cos $A = \dfrac{3}{5}$.

(ii) cosec $A = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AD}{BD}\\[1em] = \dfrac{5}{4}$

Hence, cosec $A = \dfrac{5}{4}$.

(iii) tan²A – sec²A

tan $A = \dfrac{Perpendicular}{Base}$

$= \dfrac{BD}{AD}\\[1em] = \dfrac{4}{3}$

sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AB}{AD}\\[1em] = \dfrac{5}{3}$

tan²A – sec²A

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

Hence, tan²A – sec²A = -1.

(iv) sin $C = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BD}{BC}\\[1em] = \dfrac{4}{12}\\[1em] = \dfrac{1}{3}\\[1em]$

Hence, sin $C = \dfrac{1}{3}$.

(v) sec $C = \dfrac{Hypotenuse}{Base}$

$= \dfrac{BC}{DC}\\[1em] = \dfrac{12}{8 \sqrt{2}}\\[1em] = \dfrac{3}{2 \sqrt{2}}\\[1em] = \dfrac{3 \times \sqrt{2}}{2 \sqrt{2} \times \sqrt{2}}\\[1em] = \dfrac{3 \sqrt{2}}{2 \times 2}\\[1em] = \dfrac{3 \sqrt{2}}{4}$

Hence, sec $C = \dfrac{3 \sqrt{2}}{4}$.

(vi) cot²C - $\dfrac{1}{\text{sin}^2 \text{ C}}$

cot $C = \dfrac{Base}{Perpendicular}$

$= \dfrac{Base}{Perpendicular}\\[1em] = \dfrac{8 \sqrt{2}}{4}\\[1em] = 2 \sqrt{2}$

sin $C = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{4}{12}\\[1em] = \dfrac{1}{3}$

Now, cot²C - $\dfrac{1}{\text{sin}^2 \text{ C}}$

$= (2 \sqrt{2})^2 - \dfrac{1}{\Big(\dfrac{1}{3}\Big)^2}\\[1em] = (2 \sqrt{2})^2 - \dfrac{9}{1}\\[1em] = 4 \times 2 - 9\\[1em] = 8 - 9\\[1em] = -1$

Hence, cot²C - $\dfrac{1}{\text{sin}^2 \text{ C}} = -1$.

From the following figure, find the values of :

(i) sin B

(ii) tan C

(iii) sec² B – tan² B

(iv) sin² C + cos² C

Answer

In Δ ABD,

⇒ AB² = BD² + DA² (∵ AB is hypotenuse)

⇒ 13² = 5² + DA²

⇒ 169 = 25 + DA²

⇒ DA² = 169 - 25

⇒ DA² = 144

⇒ DA = $\sqrt{144}$

⇒ DA = 12

In Δ ADC,

⇒ AC² = AD² + DC² (∵ AB is hypotenuse)

⇒ AC² = 12² + 16²

⇒ AC² = 144 + 256

⇒ AC² = 400

⇒ AC = $\sqrt{400}$

⇒ AC = 20

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

$= \dfrac{AD}{AB}\\[1em] = \dfrac{12}{13}$

Hence, sin $B = \dfrac{12}{13}$.

(ii) tan $C = \dfrac{Perpendicular}{Base}$

$= \dfrac{AD}{DC}\\[1em] = \dfrac{12}{16}\\[1em] = \dfrac{3}{4}$

Hence, tan $C = \dfrac{3}{4}$.

(iii) sec² B – tan² B

sec $B = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AB}{BD}\\[1em] = \dfrac{13}{5}$

tan $B = \dfrac{Perpendicular}{Base}$

$= \dfrac{AD}{BD}\\[1em] = \dfrac{12}{5}$

sec² B – tan² B

$= \Big(\dfrac{13}{5}\Big)^2 - \Big(\dfrac{12}{5}\Big)^2\\[1em] = \Big(\dfrac{169}{25}\Big) - \Big(\dfrac{144}{25}\Big)\\[1em] = \Big(\dfrac{169 - 144}{25}\Big)\\[1em] = \Big(\dfrac{25}{25}\Big)\\[1em] = 1$

Hence, sec² B – tan² B = 1.

(iv) sin² C + cos² C

sin $C = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{AD}{AC}\\[1em] = \dfrac{12}{20}\\[1em] = \dfrac{3}{5}$

cos $C = \dfrac{Base}{Hypotenuse}$

$= \dfrac{DC}{AC}\\[1em] = \dfrac{16}{20}\\[1em] = \dfrac{4}{5}$

Now,

sin² C + cos² C

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

Hence, sin² C + cos² C = 1.

Given : $\text{sin A} = \dfrac{3}{5}$, find :

(i) tan A

(ii) cos A

Answer

(i) Given:

sin $A = \dfrac{3}{5}$

i.e., $\dfrac{Perpendicular}{Hypotenuse} = \dfrac{3}{5}$

∴ If length of BC = 3x unit, length of AC = 5x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (5x)² = AB² + (3x)²

⇒ 25x² = AB² + 9x²

⇒ AB² = 25x² - 9x²

⇒ AB² = 16x²

⇒ AB = $\sqrt{16\text{x}^2}$

⇒ AB = 4x

tan $A = \dfrac{Perpendicular}{Base}$

= $\dfrac{BC}{BA} = \dfrac{3x}{4x} = \dfrac{3}{4}$

Hence, tan $A = \dfrac{3}{4}$.

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

$= \dfrac{BA}{AC} =\dfrac{4x}{5x} = \dfrac{4}{5}$

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

From the following figure, find the values of :

(i) sin A

(ii) sec A

(iii) cos² A + sin² A

Answer

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = a² + a²

⇒ AC² = 2a²

⇒ AC = $\sqrt{2a^2}$

⇒ AC = $a \sqrt{2}$

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

$= \dfrac{CB}{AC}\\[1em] = \dfrac{a}{a \sqrt{2}}\\[1em] = \dfrac{a \times \sqrt{2}}{a \sqrt{2} \times \sqrt{2}}\\[1em] = \dfrac{a \sqrt{2}}{a \times 2}\\[1em] = \dfrac{\cancel{a} \sqrt{2}}{\cancel{a} \times 2}\\[1em] = \dfrac{\sqrt{2}}{2}\\[1em]$

Hence, sin $A = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$.

(ii) sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB}\\[1em] = \dfrac{a \sqrt{2}}{a}\\[1em] = \dfrac{\cancel{a} \sqrt{2}}{\cancel{a}}\\[1em] = \sqrt{2}$

Hence, sec $A = \sqrt{2}$.

(iii) cos² A + sin² A

cos $A = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC}\\[1em] = \dfrac{a}{a \sqrt{2}}\\[1em] = \dfrac{a \times \sqrt{2}}{a \sqrt{2} \times \sqrt{2}}\\[1em] = \dfrac{a \sqrt{2}}{a \times 2}\\[1em] = \dfrac{\cancel{a} \sqrt{2}}{\cancel{a} \times 2}\\[1em] = \dfrac{\sqrt{2}}{2}\\[1em]$

sin $A = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{CB}{AC}\\[1em] = \dfrac{a}{a \sqrt{2}}\\[1em] = \dfrac{a \times \sqrt{2}}{a \sqrt{2} \times \sqrt{2}}\\[1em] = \dfrac{a \sqrt{2}}{a \times 2}\\[1em] = \dfrac{\cancel{a} \sqrt{2}}{\cancel{a} \times 2}\\[1em] = \dfrac{\sqrt{2}}{2}\\[1em]$

Now, cos² A + sin² A

$= \Big(\dfrac{\sqrt{2}}{2}\Big)^2 + \Big(\dfrac{\sqrt{2}}{2}\Big)^2\\[1em] = \dfrac{2}{4} + \dfrac{2}{4}\\[1em] = \dfrac{2 + 2}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

Hence, cos² A + sin² A = 1.

Given : $\text{cos A} = \dfrac{5}{13}$

evaluate :

(i) $\dfrac{\text{sin A} - \text{cot A}}{2 \text{ tan A}}$

(ii) $\text{cot A} + \dfrac{1}{\text{cos A}}$

Answer

Given:

cos $A = \dfrac{5}{13}$

i.e. $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{5}{13}$

∴ If length of BA = 5x unit, length of AC = 13x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (13x)² = (5x)² + BC²

⇒ 169x² = 25x² + BC²

⇒ BC² = 169x² - 25x²

⇒ BC² = 144x²

⇒ BC = $\sqrt{144\text{x}^2}$

⇒ BC = 12x

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

$= \dfrac{BC}{AC} = \dfrac{12x}{13x} = \dfrac{12}{13}$

cot $A = \dfrac{Base}{Perpendicular}$

$= \dfrac{AB}{BC} = \dfrac{5x}{12x} = \dfrac{5}{12}$

tan $A = \dfrac{Perpendicular}{Base}$

$= \dfrac{BC}{AB} = \dfrac{12x}{5x} = \dfrac{12}{5}$

Now,

$= \dfrac{\text{sin A} - \text{cot A}}{2 \text{ tan A}}\\[1em] = \dfrac{\dfrac{12}{13} - \dfrac{5}{12}}{2 \times \dfrac{12}{5}}\\[1em] = \dfrac{\dfrac{12 \times 12}{13 \times 12} - \dfrac{5 \times 13}{12 \times 13}}{\dfrac{24}{5}}\\[1em] = \dfrac{\dfrac{144}{156} - \dfrac{65}{156}}{\dfrac{24}{5}}\\[1em] = \dfrac{\dfrac{144 - 65}{156}}{\dfrac{24}{5}}\\[1em] = \dfrac{\dfrac{79}{156}}{\dfrac{24}{5}}\\[1em] = \dfrac{79 \times 5}{156 \times 24}\\[1em] = \dfrac{395}{3,744}$

Hence, $\dfrac{\text{sin A} - \text{cot A}}{2 \text{ tan A}} = \dfrac{395}{3,744}$.

(ii) cos $A = \dfrac{5}{13}$

cot $A = \dfrac{5}{12}$

To find,

$\text{cot A} + \dfrac{1}{\text{cos A}}$

$\text{cot A} + \dfrac{1}{\text{cos A}} = \dfrac{5}{12} + \dfrac{1}{\dfrac{5}{13}}\\[1em] = \dfrac{5}{12} + \dfrac{13}{5}\\[1em] = \dfrac{5 \times 5}{12 \times 5} + \dfrac{13 \times 12}{5 \times 12}\\[1em] = \dfrac{25}{60} + \dfrac{156}{60}\\[1em] = \dfrac{25 + 156}{60}\\[1em] = \dfrac{181}{60}$

Hence, $\text{cot A} + \dfrac{1}{\text{cos A}} = \dfrac{181}{60}$.

Given : $\text{sec A} =\dfrac{29}{21}$, evaluate : $\text{sin A} - \dfrac{1}{\text{tan A}}$

Answer

Given:

sec $A = \dfrac{29}{21}$

i.e. $\dfrac{\text{Hypotenuse}}{\text{Base}} = \dfrac{29}{21}$

∴ If length of AB = 21x unit, length of AC = 29x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (29x)² = (21x)² + BC²

⇒ 841x² = 441x² + BC²

⇒ BC² = 841x² - 441x²

⇒ BC² = 400x²

⇒ BC = $\sqrt{400\text{x}^2}$

⇒ BC = 20x

sin $A = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BC}{AC} = \dfrac{20x}{29x} = \dfrac{20}{29}$

tan $A = \dfrac{Perpendicular}{Base}$

$= \dfrac{BC}{AB} = \dfrac{20x}{21x} = \dfrac{20}{21}$

Now,

$\text{sin A} - \dfrac{1}{\text{tan A}}\\[1em] = \dfrac{20}{29} - \dfrac{1}{\dfrac{20}{21}}\\[1em] = \dfrac{20}{29} - \dfrac{21}{20}\\[1em] = \dfrac{20 \times 20}{29 \times 20} - \dfrac{21 \times 29}{20 \times 29}\\[1em] = \dfrac{400}{580} - \dfrac{609}{580}\\[1em] = \dfrac{400 - 609}{580}\\[1em] = \dfrac{- 209}{580}\\[1em]$

Hence, $\text{sin A} - \dfrac{1}{\text{tan A}} = \dfrac{- 209}{580}$.

Given : $\text{tan A} = \dfrac{4}{3}$, find : $\dfrac{\text{cosec A}}{\text{cot A} - \text{sec A}}$

Answer

Given:

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

i.e., $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{4}{3}$

∴ If length of AB = 3x unit, length of BC = 4x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25\text{x}^2}$

⇒ AC = 5x

cosec $A = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AC}{BC} = \dfrac{5x}{4x} = \dfrac{5}{4}$

cot $A = \dfrac{Base}{Perpendicular}$

$= \dfrac{AB}{BC} = \dfrac{3x}{4x} = \dfrac{3}{4}$

sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{5x}{3x} = \dfrac{5}{3}$

Now,

$\dfrac{\text{cosec A}}{\text{cot A} - \text{sec A}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{3}{4} - \dfrac{5}{3}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{3 \times 3}{4 \times 3} - \dfrac{5 \times 4}{3 \times 4}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{9}{12} - \dfrac{20}{12}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{9 - 20}{12}}\\[1em] = \dfrac{\dfrac{5}{4}}{\dfrac{- 11}{12}}\\[1em] = \dfrac{5 \times 12}{- 11 \times 4}\\[1em] = \dfrac{- 60}{44}\\[1em] = \dfrac{- 15}{11}$

Hence, $\dfrac{\text{cosec A}}{\text{cot A} - \text{sec A}} = \dfrac{- 15}{11}$.

Given : 4 cot A = 3, find :

(i) sin A

(ii) sec A

(iii) cosec² A - cot² A

Answer

Given:

4 cot A = 3

cot $A = \dfrac{3}{4}$

i.e., $\dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{3}{4}$

∴ If length of AB = 3x unit, length of BC = 4x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25\text{x}^2}$

⇒ AC = 5x

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

$= \dfrac{BC}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$

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

(ii) sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{5x}{3x} = \dfrac{5}{3} = 1\dfrac{2}{3}$

Hence, sec $A = \dfrac{5}{3} = 1\dfrac{2}{3}$.

(iii) cosec² A - cot² A

cosec $A = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AC}{BC} = \dfrac{5x}{4x} = \dfrac{5}{4}$

cot $A = \dfrac{Base}{Perpendicular}$

$= \dfrac{AB}{BC} = \dfrac{3x}{4x} = \dfrac{3}{4}$

Now,

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

Hence, cosec² A - cot² A = 1.

Given : cos A = 0.6; find all other trigonometrical ratios for angle A.

Answer

Given:

cos A = 0.6

cos $A = \dfrac{6}{10}$

cos $A = \dfrac{3}{5}$

i.e. $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{3}{5}$

∴ If length of AB = 3x unit, length of AC = 5x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ (5x)² = (3x)² + BC²

⇒ 25x² = 9x² + BC²

⇒ BC² = 25x² - 9x²

⇒ BC² = 16x²

⇒ BC = $\sqrt{16\text{x}^2}$

⇒ BC = 4x

sin $A = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BC}{AC} = \dfrac{4x}{5x} = \dfrac{4}{5}$

tan $A = \dfrac{Perpendicular}{Base}$

$= \dfrac{BC}{AB} = \dfrac{4x}{3x} = \dfrac{4}{3} = 1\dfrac{1}{3}$

cot $A = \dfrac{Base}{Perpendicular}$

$= \dfrac{AB}{BC} = \dfrac{3x}{4x} = \dfrac{3}{4}$

cosec $A = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AC}{BC} = \dfrac{5x}{4x} = \dfrac{5}{4} = 1\dfrac{1}{4}$

sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{5x}{3x} = \dfrac{5}{3} = 1\dfrac{2}{3}$

Hence, sin $A = \dfrac{4}{5}$, tan $A = 1\dfrac{1}{3}$, cot $A = \dfrac{3}{4}$, cosec $A = 1\dfrac{1}{4}$ and sec $A = 1\dfrac{2}{3}$.

In a right-angled triangle, it is given that A is an acute angle and $\text{tan A} = \dfrac{5}{12}$.

Find the values of :

(i) cos A

(ii) sin A

(iii) $\dfrac{\text{cos A} + \text{sin A}}{\text{cos A} - \text{sin A}}$

Answer

Given:

tan $A = \dfrac{5}{12}$

i.e., $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{5}{12}$

∴ If length of BC = 5x unit, length of AB = 12x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (12x)²

⇒ AC² = 25x² + 144x²

⇒ AC² = 169x²

⇒ AC = $\sqrt{169\text{x}^2}$

⇒ AC = 13x

(i) cos $A = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC} = \dfrac{12x}{13x} = \dfrac{12}{13}$

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

(ii) sin $A = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BC}{AC} = \dfrac{5x}{13x} = \dfrac{5}{13}$

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

(iii) $\dfrac{\text{cos A} + \text{sin A}}{\text{cos A} - \text{sin A}}$

$= \dfrac{\dfrac{12}{13} + \dfrac{5}{13}}{\dfrac{12}{13} - \dfrac{5}{13}}\\[1em] = \dfrac{\dfrac{12 + 5}{13}}{\dfrac{12 - 5}{13}}\\[1em] = \dfrac{\dfrac{17}{\cancel{13}}}{\dfrac{7}{\cancel{13}}}\\[1em] = \dfrac{17}{7}\\[1em] = 2\dfrac{3}{7}$

Hence, $\dfrac{\text{cos A} + \text{sin A}}{\text{cos A} - \text{sin A}} = 2\dfrac{3}{7}$.

Given : $\text{sin θ} = \dfrac{p}{q}$, find cos θ + sin θ in terms of p and q.

Answer

Given:

sin θ = $\dfrac{p}{q}$

i.e. $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{p}{q}$

∴ If length of BC = px unit, length of AC = qx unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ (qx)² = (px)² + AB²

⇒ AB² = q²x² - p²x²

⇒ AB = $\sqrt{q^2\text{x}^2 - p^2\text{x}^2}$

⇒ AB = ($\sqrt{q^2 - p^2}$) x

cos θ = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC} = \dfrac{\sqrt{(q^2 - p^2)}x}{qx} = \dfrac{\sqrt{q^2 - p^2}}{q}$

Now,

$\text{cos θ} + \text{sin θ} = \dfrac{\sqrt{q^2 - p^2}}{q} + \dfrac{p}{q}\\[1em] = \dfrac{\sqrt{q^2 - p^2} + p}{q}$

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

If $\text{cos A} = \dfrac{1}{2}$ and $\text{sin B} = \dfrac{1}{\sqrt2}$, find the value of : $\dfrac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}}$. Here angles A and B are from different right triangles.

Answer

Given:

cos $A = \dfrac{1}{2}$

i.e., $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{1}{2}$

∴ If length of AM = 1x unit, length of AO = 2x unit.

In Δ AMO,

⇒ AO² = AM² + MO² (∵ AC is hypotenuse)

⇒ (2x)² = (1x)² + MO²

⇒ 4x² = 1x² + MO²

⇒ MO² = 4x² - 1x²

⇒ MO² = 3x²

⇒ MO = $\sqrt{3\text{x}^2}$

⇒ MO = $\sqrt{3}$ x

tan $A = \dfrac{Perpendicular}{Base}$

$= \dfrac{OM}{MA} = \dfrac{\sqrt{3} x}{1x} = \dfrac{\sqrt{3}}{1} = \sqrt{3}$

And,

sin $B = \dfrac{1}{\sqrt{2}}$

i.e., $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{1}{\sqrt{2}}$

∴ If length of XY = y unit, length of YB = y $\sqrt{2}$ unit.

In Δ BXY,

⇒ YB² = YX² + BX² (∵ AC is hypotenuse)

⇒ ($\sqrt{2}$y)² = (y)² + BX²

⇒ 2y² = y² + BX²

⇒ BX² = 2y² - y²

⇒ BX² = y²

⇒ BX = $\sqrt{\text{y}^2}$

⇒ BX = y

tan $B = \dfrac{Perpendicular}{Base}$

$= \dfrac{YX}{XB} = \dfrac{y}{y}= 1$

Now,

$\dfrac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}}\\[1em] = \dfrac{{\sqrt{3} - 1}}{1 + {\sqrt{3}}}\\[1em] = \dfrac{(\sqrt{3} - 1) \times (1 - \sqrt{3})}{(1 + \sqrt{3}) \times (1 - \sqrt{3})}\\[1em] = \dfrac{(\sqrt{3} - 3 - 1 + \sqrt{3})}{1^2 - \sqrt{3}^2}\\[1em] = \dfrac{2 \sqrt{3} - 4}{1 - 3}\\[1em] = \dfrac{2 (\sqrt{3} - 2)}{-2}\\[1em] = \dfrac{\cancel{2} (\sqrt{3} - 2)}{-\cancel{2}}\\[1em] = -\sqrt{3} + 2\\[1em] = 2 - \sqrt{3}$

Hence, $\dfrac{\text{tan A} - \text{tan B}}{1 + \text{tan A } \text{tan B}} = 2 - \sqrt{3}$.