Chapter 21: Trigonometrical Ratios

Exercise 21(A)

Question 1(a)

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

$\dfrac{5}{12}$
$\dfrac{12}{13}$
$\dfrac{12}{5}$
$\dfrac{13}{12}$

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.

If sin A = 5/13 the value of tan A is : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

In Δ ABC,

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

⇒ (13x)² = (5x)² + BC²

⇒ 169x² = 25x² + BC²

⇒ BC² = 169x² - 25x²

⇒ BC² = 144x²

⇒ BC = $\sqrt{144x^2}$

⇒ BC = 12x

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

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

Hence, option 1 is the correct option.

Question 1(b)

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

$\dfrac{9}{25}$
1
$\dfrac{9}{16}$
$\dfrac{16}{9}$

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.

If tan A = 3/5, the value of sin2 A + cos2 A is : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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.

Question 1(c)

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

1
2
-2
-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.

If cot A = 5/12, the value of cot2 A - cosec2 A is : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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.

Question 1(d)

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

$\dfrac{3}{5}$
$\dfrac{4}{3}$
$\dfrac{3}{4}$
$\dfrac{4}{5}$

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.

Question 1(e)

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

1
-1
$\dfrac{3}{4}$
$\dfrac{4}{3}$

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.

If 5 cos A = 3, the value of sec2 A - tan2 A is : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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.

Question 2

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

Question 3

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.

Question 4

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

Question 5

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.

Question 6

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.

Given : sin A = 3/5, find : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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

Question 7

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

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.

Question 8

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.

Given : cos A = 5/13. Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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

Question 9

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.

Given : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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

Question 10

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

Question 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.

Given : 4 cot A = 3, find : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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.

Question 12

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.

Given : cos A = 0.6; find all other trigonometrical ratios for angle A. Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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 $A = 1\dfrac{2}{3}$ .

Question 13

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 a right-angled triangle, it is given that A is an acute angle and tan A = 5/12. Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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

Question 14

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.

Given : sin θ = p/q, find cos θ + sin θ in terms of p and q. Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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 = $\dfrac{\sqrt{q^2 - p^2} + p}{q}$ .

Question 15

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.

Here angles A and B are from different right triangles. Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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

Exercise 21(B)

Question 1(a)

If $\text{cos A} = \dfrac{1}{\sqrt2}$ and $\text{sin B} = \dfrac{\sqrt3}{2}$ , the value of $\dfrac{\text{tan B} - \text{tan A}}{1 + \text{tan A tan B}}$ is :

$2 - {\sqrt3}$
$2 + {\sqrt3}$
${\sqrt3} - 2$
${\sqrt3} + 2$

Answer

Given:

cos $A = \dfrac{1}{\sqrt2}$

i.e., $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{1}{\sqrt2}$

∴ If length of AM = x unit, length of AO = x $\sqrt{2}$ unit.

If cos A = 1/2 and sin B = 3/2, the value of tan B - tan A1 + tan A tan B is : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

In Δ AMO,

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

⇒ ( $\sqrt{2}$ x)² = (x)² + MO²

⇒ 2x² = x² + MO²

⇒ MO² = 2x² - x²

⇒ MO² = x²

⇒ MO = $\sqrt{\text{x}^2}$

⇒ MO = x

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

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

And,

sin $B = \dfrac{\sqrt3}{2}$

i.e. $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{\sqrt3}{2}$

∴ If length of XY = $\sqrt{3}$ y unit, length of YB = 2y unit.

In Δ BXY,

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

⇒ (2y)² = ( $\sqrt{3}$ y)² + BX²

⇒ 4y² = 3y² + BX²

⇒ BX² = 4y² - 3y²

⇒ BX² = y²

⇒ BX = $\sqrt{\text{y}^2}$

⇒ BX = y

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

$= \dfrac{XY}{BX} = \dfrac{\sqrt{3}y}{y} = \dfrac{\sqrt{3}}{1}$

Now,

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

Hence, option 1 is the correct option.

Question 1(b)

From the given figure, the value of cos y is :

$\dfrac{1}{3}$
$\dfrac{1}{4}$
$\dfrac{12}{13}$
$1\dfrac{1}{12}$

Answer

In Δ ABC,

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

⇒ AC² = 4² + 3²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

In Δ ADC,

⇒ DC² = AC² + DA² (∵ DC is hypotenuse)

⇒ DC² = 5² + 12²

⇒ DC² = 25 + 144

⇒ DC² = 169

⇒ DC = $\sqrt{169}$

⇒ DC = 13

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

$= \dfrac{DA}{DC}\\[1em] = \dfrac{12}{13}$

Hence, option 3 is the correct option.

Question 1(c)

ABCD is a rhombus with diagonals BD and AC equal to 12 cm and 16 cm respectively. The value of cosec x is :

$1\dfrac{2}{3}$
$3\dfrac{1}{3}$
$\dfrac{1}{2}$
$1\dfrac{1}{3}$

Answer

The diagonals of a rhombus bisect each other at right angles.

∴ ∠AOB = 90°

In Δ AOB,

⇒ AB² = BO² + OA² (∵ AB is hypotenuse)

⇒ AB² = 6² + 8²

⇒ AB² = 36 + 64

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm

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

$= \dfrac{AB}{OB}\\[1em] = \dfrac{10}{6}\\[1em] = \dfrac{5}{3}\\[1em] = 1\dfrac{2}{3}$

Hence, option 1 is the correct option.

Question 1(d)

If sin A - cosec A = 2, the value of sin² A + cosec² A is :

Answer

Given:

sin A - cosec A = 2

Squaring both sides,

⇒ (sin A - cosec A)² = (2)²

⇒ (sin A)² + (cosec A)² - 2 x sin A x cosec A = 4

⇒ sin² A + cosec² A - 2 x $\text{sin A}$ x $\dfrac{1}{\text{sin A}}$ = 4

⇒ sin² A + cosec² A - 2 x $\cancel{\text{sin A}}$ x $\dfrac{1}{\cancel{\text{sin A}}}$ = 4

⇒ sin² A + cosec² A - 2 = 4

⇒ sin² A + cosec² A = 4 + 2

⇒ sin² A + cosec² A = 6

Hence, option 4 is the correct option.

Question 1(e)

If $\text{cot A} = {\sqrt5}$ , the value of cosec² A - sec² A is :

$\dfrac{5}{24}$
$4\dfrac{4}{5}$
5
24

Answer

Given:

$\text{cot A} = {\sqrt5}\\[1em] \text{cot A} = \dfrac{\text{Base}}{\text{Perpendicular}} = \dfrac{\sqrt5}{1}\\[1em]$

∴ If length of AB = $\sqrt{5}$ x unit, length of BC = x unit.

If cot A = 5, the value of cosec2 A - sec2 A is : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

In Δ ABC,

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

⇒ AC² = $(\sqrt{5}$ x)² + (x)²

⇒ AC² = 5x² + x²

⇒ AC² = 6x²

⇒ AC = $\sqrt{6\text{x}^2}$

⇒ AC = $\sqrt{6}$ x

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

$= \dfrac{AC}{BC} = \dfrac{\sqrt{6}\text{x}}{\text{x}} = \dfrac{\sqrt{6}}{1}$

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

$= \dfrac{AC}{AB} = \dfrac{\sqrt{6}\text{x}}{\sqrt{5}\text{x}} = \dfrac{\sqrt{6}}{\sqrt{5}}$

Now, cosec² A - sec² A

$= \Big(\dfrac{\sqrt{6}}{1}\Big)^2 - \Big(\dfrac{\sqrt{6}}{\sqrt{5}}\Big)^2\\[1em] = \dfrac{6}{1} - \dfrac{6}{5}\\[1em] = \dfrac{6 \times 5}{1 \times 5} - \dfrac{6 \times 1}{5 \times 1}\\[1em] = \dfrac{30}{5} - \dfrac{6}{5}\\[1em] = \dfrac{30 - 6}{5}\\[1em] = \dfrac{24}{5}\\[1em] = 4\dfrac{4}{5}$

Hence, option 2 is the correct option.

Question 2

From the following figure, find :

(i) y

(ii) sin x°

(iii) (sec x° - tan x°)(sec x° + tan x°)

Answer

(i) In Δ ABC,

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

⇒ 2² = y² + 1²

⇒ 4 = y² + 1

⇒ y² = 4 - 1

⇒ y² = 3

⇒ y = $\sqrt{3}$

Hence, the value of y = $\sqrt{3}$ .

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

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

Hence, sin $x° = \dfrac{\sqrt{3}}{2}$ .

(iii) (sec x° - tan x°)(sec x° + tan x°)

sec $x° = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{CB}\\[1em] = \dfrac{2}{1} = 2$

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

$= \dfrac{AB}{CB}\\[1em] = \dfrac{\sqrt{3}}{1} = 3$

Now, (sec x° - tan x°)(sec x° + tan x°)

$= (2 - \sqrt{3})(2 + \sqrt{3}) \\[1em] = (2)^2 - (\sqrt{3})^2 \\[1em] = 4 - 3 \\[1em] = 1$

Hence, (sec x° - tan x°)(sec x° + tan x°) = 1.

Question 3

Use the given figure to find :

(i) sin x°

(ii) cos y°

(iii) 3 tan x° - 2 sin y° + 4 cos y°

Answer

In Δ BCD,

⇒ BD² = BC² + CD² (∵ BD is hypotenuse)

⇒ BD² = 6² + 8²

⇒ BD² = 36 + 64

⇒ BD² = 100

⇒ BD = $\sqrt{100}$

⇒ BD = 10

In Δ ACD,

⇒ AD² = AC² + CD² (∵ AD is hypotenuse)

⇒ 17² = AC² + 8²

⇒ 289 = AC² + 64

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = $\sqrt{225}$

⇒ AC = 15

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

$= \dfrac{DC}{AD}\\[1em] = \dfrac{8}{17}$

Hence, sin $x° = \dfrac{8}{17}$ .

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

$= \dfrac{BC}{BD}\\[1em] = \dfrac{6}{10}\\[1em] = \dfrac{3}{5}$

Hence, cos $y° = \dfrac{3}{5}$ .

(iii) 3 tan x° - 2 sin y° + 4 cos y°

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

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

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

$= \dfrac{CD}{BD}\\[1em] = \dfrac{8}{10}\\[1em] = \dfrac{4}{5}$

cos $y° = \dfrac{Base}{Hypotenuse}$

$= \dfrac{BC}{BD}\\[1em] = \dfrac{6}{10}\\[1em] = \dfrac{3}{5}$

Now, 3 tan x° - 2 sin y° + 4 cos y°

$= 3 \times \dfrac{8}{15} - 2 \times \dfrac{4}{5} + 4 \times \dfrac{3}{5}\\[1em] = \dfrac{24}{15} - \dfrac{8}{5} + \dfrac{12}{5}\\[1em] = \dfrac{24}{15} + \dfrac{- 8 + 12}{5}\\[1em] = \dfrac{24}{15} + \dfrac{4}{5}\\[1em] = \dfrac{24}{15} + \dfrac{4 \times 3}{5 \times 3}\\[1em] = \dfrac{24}{15} + \dfrac{12}{15}\\[1em] = \dfrac{24 + 12}{15}\\[1em] = \dfrac{36}{15}\\[1em] = \dfrac{12}{5}\\[1em] = 2\dfrac{2}{5}$

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

Question 4

In the diagram, given below, triangle ABC is right-angled at B and BD is perpendicular to AC. Find :

(i) cos ∠DBC

(ii) cot ∠DBA

In the diagram, given below, triangle ABC is right-angled at B and BD is perpendicular to AC. Find : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

Answer

In Δ ABC,

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

⇒ AC² = 12² + 5²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$

⇒ AC = 13

Let ∠CBD be x°.

So, ∠DBA = 90° - x°.

In Δ DAB, according to angle sum property

⇒ ∠ DAB + ∠ADB + ∠DBA = 180°

⇒ ∠ DAB + 90° + (90° - x°) = 180°

⇒ ∠ DAB + 180° - x° = 180°

⇒ ∠ DAB - x° = 0

⇒ ∠ DAB = x°

From figure,

∠ DAB = ∠ CAB = x°

∴ ∠ CBD = ∠ CAB = x°

(i) cos ∠DBC = cos ∠CAB = $\dfrac{Base}{Hypotenuse}$

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

Hence, cos ∠DBC = $\dfrac{12}{13}$ .

(ii) In Δ BCD, according to angle sum property

⇒ ∠ DBC + ∠DCB + ∠CDB = 180°

⇒ ∠ DCB + x° + 90° = 180°

⇒ ∠ DCB = 180° - 90° - x°

⇒ ∠ DCB = 90° - x°

From figure,

∠ DCB = ∠ ACB = 90° - x°

∴ ∠ DBA = ∠ ACB = 90° - x°

cot ∠DBA = cot ∠ACB = $\dfrac{Base}{Perpendicular}$

$= \dfrac{BC}{AB}\\[1em] = \dfrac{5}{12}$

Hence, cot ∠DBA = $\dfrac{5}{12}$ .

Question 5

In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm. Find :

(i) tan ∠DBC

(ii) sin ∠DBA

In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm. Find : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

Answer

In Δ ABC,

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

⇒ AC² = 4² + 3²

⇒ AC² = 16 + 9

⇒ AC² = 25

⇒ AC = $\sqrt{25}$

⇒ AC = 5

Let CD = y and BD = x

In Δ BCD,

⇒ BC² = CD² + BD² (∵ BC is hypotenuse)

⇒ 3² = y² + x²

⇒ 9 = y² + x² ................(1)

In Δ ABD,

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

⇒ 4² = (5 - y)² + x²

⇒ 16 = 25 + y² - 10y + x² ................(2)

Subtracting (1) from (2), we get

⇒ 16 - 9 = (25 + y² - 10y + x²) - (y² + x²)

⇒ 7 = 25 + y² - 10y + x² - y² - x²

⇒ 7 = 25 - 10y

⇒ 7 - 25 = -10y

⇒ -10y = -18

⇒ y = $\dfrac{18}{10}$ = 1.8

AD = 5 - 1.8 = 3.2

Using equation (1),

⇒ 9 = (1.8)² + x²

⇒ x² = 9 - 3.24

⇒ x² = 5.76

⇒ x = $\sqrt{5.76}$

⇒ x = 2.4

(i) cos ∠DBC = $\dfrac{Perpendicular}{Base}$

$= \dfrac{CD}{BD}\\[1em] = \dfrac{1.8}{2.4}\\[1em] = \dfrac{3}{4}$

Hence, tan ∠DBC = $\dfrac{3}{4}$ .

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

$= \dfrac{AD}{AB}\\[1em] = \dfrac{3.2}{4}\\[1em] = \dfrac{4}{5}$

Hence, sin ∠DBA = $\dfrac{4}{5}$ .

Question 6

In triangle ABC, AB = AC = 15 cm and BC = 18 cm, find cos ∠ABC.

Answer

In isosceles triangle ABC, the perpendicular drawn from angle A to the side BC divides BC into 2 equal parts.

In triangle ABC, AB = AC = 15 cm and BC = 18 cm, find cos ∠ABC. Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

BD = DC = $\dfrac{18}{2}$ = 9

cos ∠ABC = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{BD}{AB}\\[1em] = \dfrac{9}{15}\\[1em] = \dfrac{3}{5}$

Hence, cos ∠ABC = $\dfrac{3}{5}$ .

Question 7

In the figure, given below, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm. Find :

(i) sin B

(ii) tan C.

(iii) sin² B + cos² B

(iv) tan C - cot B

In the figure, given below, ABC is an isosceles triangle with BC = 8 cm and AB = AC = 5 cm. Find : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

Answer

In isosceles Δ ABC, the perpendicular drawn from angle A to the side BC divides BC into 2 equal parts.

BD = DC = $\dfrac{8}{2}$ = 4

In Δ ABD,

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

⇒ 5² = 4² + AD²

⇒ 25 = 16 + AD²

⇒ AD² = 25 - 16

⇒ AD² = 9

⇒ AD = $\sqrt{9}$

⇒ AD = 3

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

$\dfrac{AD}{AB} = \dfrac{3}{5}$

Hence, sin B = $\dfrac{3}{5}$

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

$\dfrac{AD}{DC} = \dfrac{3}{4}$

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

(iii) sin² B + cos² B

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

$\dfrac{AD}{AB} = \dfrac{3}{5}$

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

$\dfrac{BD}{AB} = \dfrac{4}{5}$

Now, sin² B + cos² B

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

Hence, sin² B + cos² B = 1.

(iv) tan C - cot B

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

$= \dfrac{AD}{DC} = \dfrac{3}{4}$

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

$= \dfrac{BD}{AD} = \dfrac{4}{3}$

Now, tan C - cot B

$= \dfrac{3}{4} - \dfrac{4}{3}\\[1em] = \dfrac{3 \times 3}{4 \times 3}- \dfrac{4 \times 4}{3 \times 4}\\[1em] = \dfrac{9}{12} - \dfrac{16}{12}\\[1em] = \dfrac{9 - 16}{12}\\[1em] = \dfrac{-7}{12}$

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

Question 8

In triangle ABC; ∠ABC = 90°, ∠CAB = x°, $\text{tan x°} = \dfrac{3}{4}$ and BC = 15 cm. Find the measures of AB and AC.

Answer

Given:

$\text{tan x°} = \dfrac{3}{4}\\[1em] \text{tan x°} = \dfrac{Perpendicular}{Base} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{BC}{AB} = \dfrac{3}{4}$

In triangle ABC; ∠ABC = 90°, ∠CAB = x°, tan x° = 3/4 and BC = 15 cm. Find the measures of AB and AC. Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

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

BC = 15 cm (∵ Given)

∴ 3x = 15

⇒ x = $\dfrac{15}{3}$ = 5 cm

∴ AB = 4x = 4 x 5 = 20 cm

In Δ ABC,

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

⇒ AC² = (15)² + (20)²

⇒ AC² = 225 + 400

⇒ AC² = 625

⇒ AC = $\sqrt{625}$

⇒ AC = 25 cm

Hence, AB = 20 cm and AC = 25 cm.

Question 9

Using the measurements given in the following figure :

(i) Find the value of sin Φ and tan θ.

(ii) Write an expression for AD in terms of θ.

Using the measurements given in the following figure : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

Answer

(i) In Δ BCD,

⇒ BD² = BC² + CD² (∵ BD is hypotenuse)

⇒ 13² = 12² + CD²

⇒ 169 = 144 + CD²

⇒ CD² = 169 - 144

⇒ CD² = 25

⇒ CD = $\sqrt{25}$

⇒ CD = 5

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

$= \dfrac{CD}{BD} = \dfrac{5}{13}$

Draw a line parallel to BC from point D such that it meets AB at point E. This line DE will be ⊥ to AB.

From figure,

DE = BC = 12

In Δ BED,

⇒ BD² = BE² + DE² (∵ BD is hypotenuse)

⇒ 13² = BE² + 12²

⇒ 169 = BE² + 144

⇒ BE² = 169 - 144

⇒ BE² = 25

⇒ BE = $\sqrt{25}$

⇒ BE = 5

And, AE = AB - BE = 14 - 5 = 9

tan θ = $\dfrac{Base}{Perpendicular}$

$= \dfrac{DE}{AE} = \dfrac{12}{9} = \dfrac{4}{3}$

Hence, sin Φ = $\dfrac{5}{13}$ and tan θ = $\dfrac{4}{3}$

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

sin θ = $\dfrac{DE}{AD} = \dfrac{12}{AD}$

AD = $\dfrac{12}{\text{sin θ}}$ = 12 cosec θ

cos θ = $\dfrac{Base}{Hypotenuse}$

cos θ = $\dfrac{AE}{AD} = \dfrac{9}{AD}$

AD = $\dfrac{9}{\text{cos θ}}$ = 9 sec θ

Hence, AD = 12 cosec θ or 9 sec θ.

Question 10

In the given figure; BC = 15 cm and $\text{sin B} = \dfrac{4}{5}$ .

(i) Calculate the measures of AB and AC.

(ii) Now, if tan ∠ADC = 1; calculate the measures of CD and AD.

In the given figure; BC = 15 cm and sin B = 4/5. Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

Answer

(i) Given:

$\text{sin B} = \dfrac{4}{5}\\[1em] \text{sin B} = \dfrac{Perpendicular}{Hypotenuse} = \dfrac{4}{5}$

∴ If length of AC = 4x cm, length of AB = 5x cm.

In Δ ABC,

⇒ AB²= BC² + AC² (∵ AB is hypotenuse)

⇒ (5x)² = BC² + (4x)²

⇒ 25x² = BC² + 16x²

⇒ BC² = 25x² - 16x²

⇒ BC² = 9x²

⇒ BC = $\sqrt{9\text{x}^2}$

⇒ BC = 3x

It is given that BC = 15 cm

3x = 15

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

x = 5 cm

AB = 5x = 5 x 5 cm = 25 cm

AC = 4x = 4 x 5 cm = 20 cm

Hence, AB = 25 cm and AC = 20 cm.

(ii) tan ∠ADC = 1

$\text{tan ∠ADC} = \dfrac{Perpendicular}{Base} = 1$

∴ If length of AC = x unit, length of CD = x unit.

From (i), we know AC = 20 cm

∴ x = 20 cm

So, AC = CD = 20 cm

In Δ ACD,

⇒ AD² = AC² + CD² (∵ AD is hypotenuse)

⇒ AD² = 20² + 20²

⇒ AD² = 400 + 400

⇒ AD² = 800

⇒ AD = $\sqrt{800}$

⇒ AD = $20\sqrt{2}$

Hence, CD = 20 cm and AD = 20 $\sqrt{2}$ cm.

Question 11

If sin A + cosec A = 2; find the value of sin² A + cosec² A.

Answer

sin A + cosec A = 2

Squaring both sides,

(sin A + cosec A)² = 2²

⇒ sin² A + cosec² A + 2 x sin A x cosec A = 4

⇒ sin² A + cosec² A + 2 x sin A x $\dfrac{1}{\text{sin A}}$ = 4

⇒ sin² A + cosec² A + 2 x ${\cancel{\text{sin A}}}$ x $\dfrac{1}{\cancel{\text{sin A}}}$ = 4

⇒ sin² A + cosec² A + 2 = 4

⇒ sin² A + cosec² A = 4 - 2

⇒ sin² A + cosec² A = 2

Hence, sin² A + cosec² A = 2.

Question 12

If tan A + cot A = 5; find the value of tan² A + cot² A.

Answer

tan A + cot A = 5

Squaring both sides,

(tan A + cot A)² = 5²

⇒ tan² A + cot² A + 2 x tan A x cot A = 25

⇒ tan² A + cot² A + 2 x tan A x $\dfrac{1}{\text{tan A}}$ = 25

⇒ tan² A + cot² A + 2 x ${\cancel{\text{tan A}}}$ x $\dfrac{1}{\cancel{\text{tan A}}}$ = 25

⇒ tan² A + cot² A + 2 = 25

⇒ tan² A + cot² A = 25 - 2

⇒ tan² A + cot² A = 23

Hence, tan² A + cot² A = 23.

Question 13

Given : 4 sin θ = 3 cos θ; find the value of :

(i) sin θ

(ii) cos θ

(iii) cot² θ - cosec² θ

(iv) 4 cos² θ - 3 sin² θ + 2

Answer

Given:

4 sin θ = 3 cos θ

$⇒ \dfrac{\text{sin θ}}{\text{cos θ}} = \dfrac{3}{4}\\[1em] ⇒ \text{tan θ} = \dfrac{3}{4}\\[1em] ⇒ \text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}}\\[1em] ⇒ \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{3}{4}\\[1em]$

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

Given : 4 sin θ = 3 cos θ; find the value of : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

In Δ ABC,

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

⇒ AC² = (3x)² + (4x)²

⇒ AC² = 9x² + 16x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25 \text{x}^2}$

⇒ AC = 5x

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

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

Hence, sin θ = $\dfrac{3}{5}$ .

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

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

Hence, cos θ = $\dfrac{4}{5}$ .

(iii) cot² θ - cosec² θ + 2

cot θ = $\dfrac{Base}{Perpendicular}$

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

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

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

Now, cot² θ - cosec² θ

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

Hence, cot² θ - cosec² θ + 2 = -1.

(iv) 4 cos² θ - 3 sin² θ + 2

cos θ = $\dfrac{Base}{Hypotenuse}$

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

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

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

Now, 4 cos² θ - 3 sin² θ + 2

$= 4 \times \Big(\dfrac{4}{5}\Big)^2 - 3 \times \Big(\dfrac{3}{5}\Big)^2 + 2\\[1em] = 4 \times \dfrac{16}{25} - 3 \times \dfrac{9}{25} + 2\\[1em] = \dfrac{64}{25} - \dfrac{27}{25} + 2\\[1em] = \dfrac{64 - 27}{25} + 2\\[1em] = \dfrac{37}{25} + 2\\[1em] = \dfrac{37}{25} + \dfrac{2 \times 25}{25}\\[1em] = \dfrac{37}{25} + \dfrac{50}{25}\\[1em] = \dfrac{37 + 50}{25}\\[1em] = \dfrac{87}{25}\\[1em] = 3\dfrac{12}{25}$

Hence, 4 cos² θ - 3 sin² θ + 2 = $3\dfrac{12}{25}$ .

Question 14

Given : 17 cos θ = 15; find the value of tan θ + 2 sec θ.

Answer

Given:

17 cos θ = 15

cos θ = $\dfrac{15}{17}$

$\Rightarrow \text{cos θ} = \dfrac{Base}{Hypotenuse} = \dfrac{15}{17}\\[1em]$

Given : 17 cos θ = 15; find the value of tan θ + 2 sec θ. Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

∴ If length of AC = 17x unit, length of AB = 15x unit.

In Δ ABC,

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

⇒ (17x)² = BC² + (15x)²

⇒ 289x² = BC² + 225x²

⇒ BC² = 289x² - 225x²

⇒ BC = $\sqrt{64 \text{x}^2}$

⇒ BC = 8x

tan θ = $\dfrac{Perpendicular}{Base}$

$= \dfrac{CB}{AB} = \dfrac{8x}{15x} = \dfrac{8}{15}$

sec θ = $\dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{17x}{15x} = \dfrac{17}{15}$

Now, tan θ + 2 sec θ

$= \dfrac{8}{15} + 2 \times \dfrac{17}{15}\\[1em] = \dfrac{8}{15} + \dfrac{34}{15}\\[1em] = \dfrac{8 + 34}{15}\\[1em] = \dfrac{42}{15}\\[1em] = \dfrac{14}{5}\\[1em] = 2\dfrac{4}{5}$

Hence, tan θ + 2 sec θ = $2\dfrac{4}{5}$ .

Question 15

Given: 5 cos A - 12 sin A = 0; evaluate :

$\dfrac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}}$

Answer

Given:

5 cos A - 12 sin A = 0

⇒ 5 cos A = 12 sin A

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

⇒ $\text{tan A} = \dfrac{5}{12}$

$⇒ \text{tan A} = \dfrac{Perpendicular}{Base} = \dfrac{5}{12} \\[1em]$

Given: 5 cos A - 12 sin A = 0; evaluate : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

∴ 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

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

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

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

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

Now,

$\dfrac{\text{sin A} + \text{cos A }}{2\text{ cos A } - \text{sin A}}\\[1em] = \dfrac{\dfrac{5}{13} + \dfrac{12}{13}}{2 \times \dfrac{12}{13} - \dfrac{5}{13}}\\[1em] = \dfrac{\dfrac{5 + 12}{13}}{\dfrac{24}{13} - \dfrac{5}{13}}\\[1em] = \dfrac{\dfrac{17}{13}}{\dfrac{24 - 5}{13}}\\[1em] = \dfrac{\dfrac{17}{13}}{\dfrac{19}{13}}\\[1em] = \dfrac{\dfrac{17}{\cancel{13}}}{\dfrac{19}{\cancel{13}}}\\[1em] = \dfrac{17}{19}$

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

Exercise 21(C)

Question 1(a)

$\dfrac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$ is equal to :

sin 60°
cos 60°
sec 60°
cosec 60°

Answer

$\dfrac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}} = \dfrac{2 \times \dfrac{1}{\sqrt{3}}}{1 + \Big(\dfrac{1}{\sqrt{3}}\Big)^2}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{2 \times 3}{4 \times {\sqrt{3}}}\\[1em] = \dfrac{6}{4{\sqrt{3}}}\\[1em] = \dfrac{3}{2{\sqrt{3}}}\\[1em] = \dfrac{{\sqrt{3}}}{2}\\[1em] = \text{sin} 60°$

Hence, option 1 is the correct option.

Question 1(b)

If tan 3A - ${\sqrt3}$ = 0 and 0 ≤ 3A ≤ 90° ; the measure of angle A is :

15°
20°
30°
10°

Answer

tan 3A - ${\sqrt3}$ = 0

tan 3A = ${\sqrt3}$

tan 3A = tan 60°

So, 3A = 60°

A = $\dfrac{60°}{3}$

A = 20°

Hence, option 2 is the correct option.

Question 1(c)

If cot A = tan A and 0 ≤ A ≤ 90°, the measure of angle A is :

30°
60°
45°
90°

Answer

cot A = tan A

⇒ $\dfrac{1}{\text{tan A}}$ = tan A

⇒ tan² A = 1

⇒ tan A = 1

⇒ tan A = tan 45°

Hence, A = 45°

Hence, option 3 is the correct option.

Question 1(d)

The value of :

cos² 60° - 2 sin³ 30° + 3 cot⁴ 45° is :

1
-2
3
2

Answer

$\text{cos}^2 60° - 2 \text{sin}^3 30° + 3 \text{cot}^4 45° = \Big(\dfrac{1}{2}\Big)^2 - 2 \times \Big(\dfrac{1}{2}\Big)^3 + 3 \times (1)^2\\[1em] = \dfrac{1}{4} - 2 \times \dfrac{1}{8} + 3\\[1em] = \dfrac{1}{4} - \dfrac{1}{4} + 3\\[1em] = 3$

Hence, option 3 is the correct option.

Question 1(e)

The value of $\dfrac{\text{cos 60° - cos 0°} + \text{2 sin 90°}}{\text{ cot 60°}\times \text{ cot 30°}}$ is :

$1\dfrac{1}{2}$
$\dfrac{2}{3}$
-2
1

Prove that :

$\Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

Answer

$\Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

$\text{L.H.S.} = \Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2\\[1em] = \Big(\dfrac{\sqrt3 + 1}{\sqrt3 - 1}\Big)^2\\[1em] = \Big(\dfrac{(\sqrt3 + 1) \times (\sqrt3 + 1)}{(\sqrt3 - 1) \times (\sqrt3 + 1)}\Big)^2\\[1em] = \Big(\dfrac{(\sqrt3 + 1)^2}{(\sqrt3)^2 - (1)^2}\Big)^2\\[1em] = \Big(\dfrac{3 + 1 + 2 \times 1 \times \sqrt3}{3 - 1}\Big)^2\\[1em] = \Big(\dfrac{4 + 2\sqrt3}{2}\Big)^2\\[1em] = (2 + \sqrt3)^2\\[1em] = 4 + 3 + 2 \times 2 \times \sqrt3\\[1em] = 7 + 4\sqrt3$

$\text{R.H.S.} = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}\\[1em] = \dfrac{1 + \dfrac{\sqrt3}{2}}{1 - \dfrac{\sqrt3}{2}}\\[1em] = \dfrac{\dfrac{2 + \sqrt3}{2}}{\dfrac{2 - \sqrt3}{2}}\\[1em] = \dfrac{\dfrac{2 + \sqrt3}{\cancel{2}}}{\dfrac{2 - \sqrt3}{\cancel{2}}}\\[1em] = \dfrac{2 + \sqrt3}{2 - \sqrt3}\\[1em] = \dfrac{(2 + \sqrt3) \times (2 + \sqrt3)}{(2 - \sqrt3) \times (2 + \sqrt3)}\\[1em] = \dfrac{(2 + \sqrt3)^2}{(2)^2 - (\sqrt3)^2}\\[1em] = \dfrac{4 + 3 + 2 \times 2 \times \sqrt3}{4 - 3}\\[1em] = 7 + 4\sqrt3\\[1em]$

∴ L.H.S. = R.H.S.

Hence, $\Big(\dfrac{\text{tan 60° + 1}}{\text{tan 60° - 1}}\Big)^2 = \dfrac{\text{1 + cos 30°}}{\text{1 - cos 30°}}$

Question 4(vi)

Prove that :

3 cosec² 60° - 2 cot² 30° + sec² 45° = 0.

Answer

3 cosec² 60° - 2 cot² 30° + sec² 45° = 0.

L.H.S. = 3 cosec² 60° - 2 cot² 30° + sec² 45°

$= 3 \times \Big(\dfrac{2}{\sqrt3}\Big)^2 - 2 \times ({\sqrt3})^2 + ({\sqrt2})^2\\[1em] = 3 \times \Big(\dfrac{4}{3}\Big) - 2 \times 3 + 2\\[1em] = 4 - 6 + 2\\[1em] = 0$

R.H.S. = 0

∴ L.H.S. = R.H.S.

Hence, 3 cosec² 60° - 2 cot² 30° + sec² 45° = 0.

Question 5(i)

Prove that :

$\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

Answer

$\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

L.H.S. = sin (2 x 30°) = sin 60° = $\dfrac{\sqrt3}{2}$

R.H.S.

$= \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{2 \times \dfrac{1}{\sqrt3}}{1 + \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{4}{3}}\\[1em] = \dfrac{2 \times 3}{4 \times \sqrt3}\\[1em] = \dfrac{\sqrt3}{2}\\[1em]$

∴ L.H.S. = R.H.S.

Hence, $\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}$

Question 5(ii)

Prove that :

$\text{cos }(2 \times 30°) = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}$

Answer

$\text{cos }(2 \times 30°) = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}$

L.H.S. = cos (2 x 30°) = cos 60° = $\dfrac{1}{2}$

R.H.S.

$= \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{1 - \Big(\dfrac{1}{\sqrt3}\Big)^2}{1 + \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{1 - \dfrac{1}{3}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{3}{3} - \dfrac{1}{3}}{\dfrac{3}{3} + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{3 - 1}{3}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{\dfrac{2}{3}}{\dfrac{4}{3}}\\[1em] = \dfrac{\dfrac{2}{\cancel{3}}}{\dfrac{4}{\cancel{3}}}\\[1em] = \dfrac{2}{4}\\[1em] = \dfrac{1}{2}$

∴ L.H.S. = R.H.S.

Hence, $\text{cos }(2 \times 30°) = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}$

Question 5(iii)

Prove that :

$\text{tan} (2 \times 30°) = \dfrac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}$

Answer

$\text{tan} (2 \times 30°) = \dfrac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}$

L.H.S. = tan 2 x 30° = tan 60° = $\sqrt3$

R.H.S.

$= \dfrac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{2 \times \dfrac{1}{\sqrt3}}{1 - \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{1 - \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{3}{3} - \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{3 - 1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{2}{3}}\\[1em] = \dfrac{2 \times 3}{2 \times \sqrt3}\\[1em] = \sqrt3$

∴ L.H.S. = R.H.S.

Hence, $\text{tan} (2 \times 30°) = \dfrac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}$

Question 6

ABC is an isosceles right-angled triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios :

(i) sin 45°

(ii) cos 45°

(iii) tan 45°

ABC is an isosceles right-angled triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios : Trigonometrical Ratios of Standard Angles, Concise Mathematics Solutions ICSE Class 9.

Answer

In Δ ABC,

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

⇒ AC² = x² + x²

⇒ AC² = 2x²

⇒ AC = $\sqrt{2\text{x}^2}$

⇒ AC = $\sqrt{2}$ x

As Δ ABC is isosceles right angled triangle, ∠BAC = ∠BCA = $\dfrac{90°}{2}$ = 45°

(i) sin 45°

sin 45° = sin A = sin C

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

= $\dfrac{BC}{AC} = \dfrac{x}{\sqrt2x} = \dfrac{1}{\sqrt2}$

Hence, sin 45° = $\dfrac{1}{\sqrt2}$ .

(ii) cos 45°

cos 45° = cos A = cos C

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

= $\dfrac{AB}{AC} = \dfrac{x}{\sqrt2x} = \dfrac{1}{\sqrt2}$

Hence, cos 45° = $\dfrac{1}{\sqrt2}$ .

(iii) tan 45°

tan 45° = tan A = tan C

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

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

Hence, tan 45° = 1.

Question 7(i)

Prove that :

sin 60° = 2 sin 30° cos 30°.

Answer

sin 60° = 2 sin 30° cos 30°.

L.H.S. = sin 60° = $\dfrac{\sqrt3}{2}$

R.H.S.= 2 sin 30° cos 30°

$= 2 \times \dfrac{1}{2} \times \dfrac{\sqrt3}{2}\\[1em] = \cancel{2} \times \dfrac{1}{\cancel{2}} \times \dfrac{\sqrt3}{2}\\[1em] = \dfrac{\sqrt3}{2}$

∴ L.H.S. = R.H.S.

Hence proved, sin 60° = 2 sin 30° cos 30°.

Question 7(ii)

Prove that :

4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°) = 2

Answer

4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°) = 2

L.H.S. = 4 (sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°)

$= 4 \times \Big(\Big(\dfrac{1}{2}\Big)^4 + \Big(\dfrac{1}{2}\Big)^4\Big) - 3 \times \Big(\Big(\dfrac{1}{\sqrt2}\Big)^2 - (1)^2\Big)\\[1em] = 4 \times \Big(\dfrac{1}{16} + \dfrac{1}{16}\Big) - 3 \times \Big(\dfrac{1}{2} - 1\Big)\\[1em] = 4 \times \Big(\dfrac{1 + 1}{16}\Big) - 3 \times \Big(\dfrac{1}{2} - \dfrac{2}{2}\Big)\\[1em] = 4 \times \Big(\dfrac{2}{16}\Big) - 3 \times \Big(\dfrac{1 - 2}{2}\Big)\\[1em] = 4 \times \Big(\dfrac{1}{8}\Big) - 3 \times \Big(\dfrac{-1}{2}\Big)\\[1em] = \dfrac{1}{2} + \dfrac{3}{2}\\[1em] = \dfrac{1 + 3}{2}\\[1em] = \dfrac{ 4}{2}\\[1em] = 2$

R.H.S. = 2

∴ L.H.S. = R.H.S.

If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.

Answer

sin x = cos y

cos y = sin (90° - y)

So, sin x = sin (90° - y)

⇒ x = 90° - y

⇒ x + y = 90°

Hence, the relation between x and y is x + y = 90°.

Question 9(i)

If sin x = cos y, then x + y = 45°; write true or false.

Answer

False

Reason:

sin x = cos y

It is only possible only when x = y = 45°.

sin 45° = cos 45° = 1

Question 1(a)

If A = 30°, then $\dfrac{\text{sin 2 A}}{\text{1 - cos 2 A}}$ is equal to :

cot A
tan A
sec A
cosec A

Answer

$\dfrac{\text{sin 2 A}}{\text{1 - cos 2 A}}\\[1em] = \dfrac{\text{sin (2 x 30°)}}{\text{1 - cos (2 x 30°)}}\\[1em] = \dfrac{\text{sin 60°}}{\text{1 - cos 60°}}\\[1em] = \dfrac{\dfrac{\sqrt3}{2}}{1 - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{\sqrt3}{2}}{\dfrac{2}{2} - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{\sqrt3}{2}}{\dfrac{2 - 1}{2}}\\[1em] = \dfrac{\dfrac{\sqrt3}{2}}{\dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{\sqrt3}{\cancel{2}}}{\dfrac{1}{\cancel{2}}}\\[1em] = \dfrac{\sqrt3}{1}\\[1em] = \text {cot A}$

Hence, option 1 is the correct option.

Question 1(b)

If A = 60° and B = 30°; the value of sin A cos B + cos A sin B is equal to :

$\dfrac{1}{2}$
1
2
${\sqrt2}$

Answer

sin A cos B + cos A sin B = sin 60°. cos 30° + cos 60°. sin 30°

$= \dfrac{\sqrt3}{2} \times \dfrac{\sqrt3}{2} + \dfrac{1}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{3}{4} + \dfrac{1}{4}\\[1em] = \dfrac{3 + 1}{4} \\[1em] = \dfrac{4}{4} \\[1em] = 1$

Hence, option 2 is the correct option.

Question 1(c)

If A = 30° ; 3 sin A - 4 sin³ A is equal to:

cos 3A
tan 3A
sin 3A
cot 3A

Answer

3 sin A - 4 sin³ A = 3 sin 30° - 4 sin³ 30°

$= 3 \times \dfrac{1}{2} - 4 \times \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{3}{2} - 4 \times \dfrac{1}{8}\\[1em] = \dfrac{3}{2} - \dfrac{1}{2}\\[1em] = \dfrac{3 - 1}{2} \\[1em] = \dfrac{2}{2} \\[1em] = 1\\[1em] = \text{sin 90°}\\[1em] = \text{sin 3 x 30°}\\[1em] = \text{sin 3A}\\[1em]$

Hence, option 3 is the correct option.

Question 1(d)

If A = 30°, then cos⁴ A - sin⁴ A is equal to :

sin 60°
tan 60°
cot 60°
cos 60°

Answer

cos⁴ A - sin⁴ A = cos⁴ 30° - sin⁴ 30°

$= \Big(\dfrac{\sqrt3}{2}\Big)^4 - \Big(\dfrac{1}{2}\Big)^4\\[1em] = \dfrac{9}{16} - \dfrac{1}{16}\\[1em] = \dfrac{9 - 1}{16}\\[1em] = \dfrac{8}{16}\\[1em] = \dfrac{1}{2}\\[1em] = \text {cos 60°}$

Hence, option 4 is the correct option.

Question 1(e)

If A = 45° ; then 2 sin A cos A is equal to :

1
0
-1
2

Answer

2 sin A cos A = 2 sin 45°. cos 45°

$= 2 \times \dfrac{1}{\sqrt2} \times \dfrac{1}{\sqrt2}\\[1em] = 2 \times \dfrac{1}{2}\\[1em] = \cancel{2} \times \dfrac{1}{\cancel{2}}\\[1em] = 1$

Hence, option 1 is the correct option.

Question 2(i)

Given A = 60° and B = 30°, prove that :

sin (A + B) = sin A cos B + cos A sin B

Answer

sin (A + B) = sin A cos B + cos A sin B

L.H.S. = sin (A + B) = sin (60° + 30°)

= sin 90° = 1

R.H.S. = sin A cos B + cos A sin B

= sin 60° cos 30° + cos 60° sin 30°

$= \dfrac{\sqrt3}{2} \times \dfrac{\sqrt3}{2} + \dfrac{1}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{3}{4} + \dfrac{1}{4}\\[1em] = \dfrac{3 + 1}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

∴ L.H.S. = R.H.S.

Hence, sin (A + B) = sin A cos B + cos A sin B.

Question 2(ii)

Given A = 60° and B = 30°, prove that :

cos (A + B) = cos A cos B - sin A sin B

Answer

cos (A + B) = cos A cos B - sin A sin B

L.H.S. = cos (A + B) = cos (60° + 30°)

= cos 90° = 0

R.H.S. = cos A cos B - sin A sin B

= cos 60° cos 30° - sin 60° sin 30°

$= \dfrac{1}{2} \times \dfrac{\sqrt3}{2} - \dfrac{\sqrt3}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{\sqrt3}{4} - \dfrac{\sqrt3}{4}\\[1em] = 0$

∴ L.H.S. = R.H.S.

Hence, cos (A + B) = cos A cos B - sin A sin B.

Question 2(iii)

Given A = 60° and B = 30°, prove that :

cos (A - B) = cos A cos B + sin A sin B

Answer

cos (A - B) = cos A cos B + sin A sin B

L.H.S. = cos (A - B) = cos (60° - 30°)

= cos 30° = $\dfrac{\sqrt3}{2}$

R.H.S. = cos A cos B + sin A sin B

= cos 60° cos 30° + sin 60° sin 30°

$= \dfrac{1}{2} \times \dfrac{\sqrt3}{2} + \dfrac{\sqrt3}{2} \times \dfrac{1}{2}\\[1em] = \dfrac{\sqrt3}{4} + \dfrac{\sqrt3}{4}\\[1em] = 2 \times \dfrac{\sqrt3}{4}\\[1em] = \dfrac{\sqrt3}{2}\\[1em]$

∴ L.H.S. = R.H.S.

Hence, cos (A - B) = cos A cos B + sin A sin B.

Question 2(iv)

Given A = 60° and B = 30°, prove that :

$\text{tan (A - B)} = \dfrac{\text{tan A - tan B}}{\text{1 + \text{ tan A }} . \text{ tan B}}$

Answer

$\text{tan (A - B)} = \dfrac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}$

L.H.S. = tan (A - B) = tan (60° - 30°)

= tan 30° = $\dfrac{1}{\sqrt3}$

$\text{R.H.S.} = \dfrac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}\\[1em] = \dfrac{\text{tan 60° - tan 30°}}{\text{1 + tan} \text{ 60° } . \text{ tan 30°}}\\[1em] = \dfrac{\sqrt3 - \dfrac{1}{\sqrt3}}{1 + \sqrt3 \times \dfrac{1}{\sqrt3}}\\[1em] = \dfrac{\dfrac{\sqrt3 \times \sqrt3}{\sqrt3} - \dfrac{1}{\sqrt3}}{1 + \cancel{\sqrt3} \times \dfrac{1}{\cancel{\sqrt3}}}\\[1em] = \dfrac{\dfrac{3}{\sqrt3} - \dfrac{1}{\sqrt3}}{1 + 1}\\[1em] = \dfrac{\dfrac{3 - 1}{\sqrt3}}{1 + 1}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{2}\\[1em] = \dfrac{\dfrac{\cancel{2}}{\sqrt3}}{\cancel{2}}\\[1em] = \dfrac{1}{\sqrt3}$

∴ L.H.S. = R.H.S.

Hence, $\text{tan (A - B)} = \dfrac{\text{tan A - tan B}}{\text{1 + tan} \text{ A } . \text{ tan B}}$ .

Question 3(i)

If A = 30°, then prove that :

$\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$

Answer

$\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$

$\text{1st term} = \text{sin 2 A} = \text{sin (2 x 30°)} = \text{sin 60°} = \dfrac{\sqrt3}{2}$

$\text{2nd term} = \text{2 sin A cos A}\\[1em] = 2 \times \text{sin 30°} \times \text{cos 30°}\\[1em] = 2 \times \dfrac{1}{2} \times \dfrac{\sqrt3}{2}\\[1em] = \cancel{2} \times \dfrac{1}{\cancel{2}} \times \dfrac{\sqrt3}{2}\\[1em] = \dfrac{\sqrt3}{2}$

$\text{3rd term} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}\\[1em] = \dfrac{\text{2 tan 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{2 \times \dfrac{1}{\sqrt3}}{1 + \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{3}{3} + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{\dfrac{2}{\sqrt3}}{\dfrac{4}{3}}\\[1em] = \dfrac{2 \times 3}{4 \times \sqrt3}\\[1em] = \dfrac{\sqrt3}{2}$

∴ 1st term = 2nd term = 3rd term

Hence, $\text{sin 2 A} = \text{2 sin A cos A} = \dfrac{\text{2 tan A}}{1 + \text{tan}^2 \text{ A}}$ .

Question 3(ii)

If A = 30°, then prove that :

$\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}$

Answer

$\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}$

$\text{1st term} = \text{cos 2 A} = \text{cos (2 x 30°)} = \text{cos 60°} = \dfrac{1}{2}$

$\text{2nd term} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A}\\[1em] = \text{cos}^2 \text{ 30°} - \text{sin}^2 \text{ 30°}\\[1em] = \Big(\dfrac{\sqrt3}{2}\Big)^2 - \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{3}{4} - \dfrac{1}{4}\\[1em] = \dfrac{3 - 1}{4}\\[1em] = \dfrac{2}{4}\\[1em] = \dfrac{1}{2}$

$\text{3rd term} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}\\[1em] = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{1 - \Big(\dfrac{1}{\sqrt3}\Big)^2}{1 + \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{\dfrac{3}{3} - \dfrac{1}{3}}{\dfrac{3}{3} + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{3 - 1}{3}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{\dfrac{2}{3}}{\dfrac{4}{3}}\\[1em] = \dfrac{\dfrac{2}{\cancel{3}}}{\dfrac{4}{\cancel{3}}}\\[1em] = \dfrac{2}{4}\\[1em] = \dfrac{1}{2}$

∴ 1st term = 2nd term = 3rd term

Hence, $\text{cos 2 A} = \text{cos}^2 \text{ A} - \text{sin}^2 \text{ A} = \dfrac{1 - \text{tan}^2 \text{ A}}{1 + \text{tan}^2 \text{ A}}$

Question 3(iii)

If A = 30°, then prove that :

2 cos² A - 1 = 1 - 2 sin² A

Answer

2 cos² A - 1 = 1 - 2 sin² A

$\text{L.H.S.} = 2 \text{cos}^2 A - 1\\[1em] = 2 \text{cos}^2 30° - 1\\[1em] = 2 \times \Big(\dfrac{\sqrt3}{2}\Big)^2 - 1\\[1em] = 2 \times \dfrac{3}{4} - 1\\[1em] = \dfrac{3}{2} - \dfrac{2}{2}\\[1em] = \dfrac{3 - 2}{2}\\[1em] = \dfrac{1}{2}$

$\text{R.H.S.} = 1 - 2 \text{sin}^2 A \\[1em] = 1 - 2 \text{sin}^2 30°\\[1em] = 1 - 2 \times \Big(\dfrac{1}{2}\Big)^2\\[1em] = 1 - 2 \times \dfrac{1}{4}\\[1em] = \dfrac{2}{2} - \dfrac{1}{2}\\[1em] = \dfrac{2 - 1}{2}\\[1em] = \dfrac{1}{2}$

∴ L.H.S. = R.H.S.

Question 1(a)

If 2sin A - 1 = 0 and A is an acute angle, the measure of angle A is :

30°
45°
60°
90°

Answer

2sin A - 1 = 0

⇒ 2sin A = 1

⇒ sin A = $\dfrac{1}{2}$

⇒ sin A = sin 30°

⇒ A = 30°

Hence, option 1 is the correct option.

Question 1(b)

If cos A (cos A - 1) = 0; the measure of angle A is :

90° or 0°
90° and 0°
45° or 90°
45° and 90°

Answer

cos A (cos A - 1) = 0

⇒ cos A = 0 or cos A = 1

⇒ cos A = cos 90° or cos A = cos 0°

The measure of angle A is 90° or 0°.

Hence, option 1 is the correct option.

Question 1(c)

If tan⁴ A - 1 = 0 and angle A is acute, then A is :

30°
45°
± 45°
± 30°

Answer

tan⁴ A - 1 = 0

⇒ tan⁴A = 1

⇒ (tan A)⁴ = 1

⇒ tan A = tan 45°

⇒ A = 45°

Hence, option 2 is the correct option.

Question 1(d)

If 2sin 3A - 1 = 0 ; the value of angle A is :

20°
60°
10°
30°

Answer

2sin 3A - 1 = 0

⇒ 2sin 3A = 1

⇒ sin 3A = $\dfrac{1}{2}$

⇒ sin 3A = sin 30°

So, 3A = 30°

⇒ A = $\dfrac{30°}{3}$

⇒ A = 10°

Hence, option 3 is the correct option.

Question 1(e)

If 3tan² A - 1 = 0 and angle A is acute, the measure of angle A is :

20°
45°
60°
30°

Answer

3tan² A - 1 = 0

⇒ 3tan² A = 1

⇒ tan² A = $\dfrac{1}{3}$

⇒ tan A = $\sqrt{\dfrac{1}{3}}$

⇒ tan A = $\dfrac{1}{\sqrt3}$

⇒ tan A = tan 30°

⇒ A = 30°

Hence, option 4 is the correct option.

Question 2(i)

Solve the following equations for A, if :

2 sin A = 1

Answer

2 sin A = 1

⇒ sin A = $\dfrac{1}{2}$

⇒ sin A = sin 30°

Hence, A = 30°.

Question 2(ii)

Solve the following equations for A, if :

2 cos 2 A = 1

Answer

2 cos 2 A = 1

⇒ cos 2A = $\dfrac{1}{2}$

⇒ cos 2A = cos 60°

So, 2A = 60°

⇒ A = $\dfrac{60°}{2}$

⇒ A = 30°

Hence, A = 30°.

Question 2(iii)

Solve the following equations for A, if :

sin 3 A = $\dfrac{\sqrt3}{2}$

Answer

sin 3 A = $\dfrac{\sqrt3}{2}$

⇒ sin 3A = sin 60°

So, 3A = 60°

⇒ A = $\dfrac{60°}{3}$

⇒ A = 20°

Hence, A = 20°.

Question 2(iv)

Solve the following equations for A, if :

sec 2 A = 2

Answer

sec 2 A = 2

⇒ sec 2A = sec 60°

So, 2A = 60°

⇒ A = $\dfrac{60°}{2}$

⇒ A = 30°

Hence, A = 30°.

Question 2(v)

Solve the following equations for A, if :

${\sqrt3}$ tan A = 1

Answer

${\sqrt3}$ tan A = 1

⇒ tan A = $\dfrac{1}{\sqrt3}$

⇒ tan A = tan 30°

So, A = 30°

Hence, A = 30°.

Calculate the value of A, if :

(sin A - 1) (2 cos A - 1) = 0

Answer

(sin A - 1) (2 cos A - 1) = 0

⇒ (sin A - 1) = 0 and (2 cos A - 1) = 0

⇒ sin A = 1 and cos A = $\dfrac{1}{2}$

⇒ sin A = sin 90° and cos A = cos 60°

Hence, A = 90° and 60°.

Question 3(ii)

Calculate the value of A, if :

(tan A - 1) (cosec 3A - 1) = 0

Answer

(tan A - 1) (cosec 3A - 1) = 0

⇒ (tan A - 1) = 0 and (cosec 3A - 1) = 0

⇒ tan A = 1 and cosec 3A = 1

⇒ tan A = tan 45° and cosec 3A = cosec 90°

So, A = 45° and 3A = 90°

⇒ A = 45° and A = 30°

Hence, A = 45° and 30°.

Question 3(iii)

Calculate the value of A, if :

(sec 2A - 1) (cosec 3A - 1) = 0

Answer

(sec 2A - 1) (cosec 3A - 1) = 0

⇒ sec 2A - 1 = 0 and cosec 3A - 1 = 0

⇒ sec 2A = 1 and cosec 3A = 1

⇒ sec 2A = sec 0° and cosec 3A = cosec 90°

So, 2A = 0° and 3A = 90°

⇒ A = 0° and A = $\dfrac{90°}{3} = 30°$

Hence, A = 0° and 30°.

Question 3(iv)

Calculate the value of A, if :

cos 3A (2 sin 2A - 1) = 0

Answer

cos 3A. (2 sin 2A - 1) = 0

⇒ cos 3A = 0 and 2sin 2A - 1 = 0

⇒ cos 3A = 0 and 2sin 2A = 1

⇒ cos 3A = 0 and sin 2A = $\dfrac{1}{2}$

⇒ cos 3A = cos 90° and sin 2A = sin 30°

So, 3A = 90° and 2A = 30°

⇒ A = $\dfrac{90°}{3} = 30°$ and A = $\dfrac{30°}{2} = 15°$

Hence, A = 30° and 15°.

Question 3(v)

Calculate the value of A, if :

(cosec 2A - 2) (cot 3A - 1) = 0

Answer

(cosec 2A - 2) (cot 3A - 1) = 0

⇒ cosec 2A - 2 = 0 and cot 3A - 1 = 0

⇒ cosec 2A = 2 and cot 3A = 1

⇒ cosec 2A = cosec 30° and cot 3A = cot 45°

So, 2A = 30° and 3A = 45°

⇒ A = $\dfrac{30°}{2} = 15°$ and A = $\dfrac{45°}{3} = 15°$

Hence, A = 15°.

Question 4

If 2 sin x° - 1 = 0 and x° is an acute angle; find:

(i) sin x°

(ii) x°

(iii) cos x° and

(iv) tan x°.

Answer

(i) 2 sin x° - 1 = 0

⇒ 2 sin x° = 1

⇒ sin x° = $\dfrac{1}{2}$

Hence, sin x° = $\dfrac{1}{2}$ .

(ii) x°

⇒ sin x° = $\dfrac{1}{2}$

⇒ sin x° = sin 30°

Hence, x° = 30°.

(iii) cos x°

⇒ cos 30° = $\dfrac{\sqrt3}{2}$

Hence, cos x° = $\dfrac{\sqrt3}{2}$ .

(iv) tan x°.

⇒ tan 30° = $\dfrac{1}{\sqrt3}$

Hence, tan x° = $\dfrac{1}{\sqrt3}$ .

Question 5

If 4 cos² x° - 1 = 0 and 0 ≤ x° ≤ 90°, find:

(i) x°

(ii) sin² x° + cos² x°

(iii) $\dfrac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°}$

Answer

(i) 4 cos² x° - 1 = 0

⇒ 4 cos² x° = 1

⇒ cos² x° = $\dfrac{1}{4}$

⇒ cos x° = $\sqrt{\dfrac{1}{4}}$

⇒ cos x° = $\dfrac{1}{2}$

⇒ cos x° = cos 60°

Hence, x° = 60°.

(ii) sin² x° + cos² x°

⇒ sin² 60° + cos² 60°

$= \Big(\dfrac{\sqrt3}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{3 + 1}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

Hence, sin² x° + cos² x° = 1.

(iii) $\dfrac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°}$

$= \dfrac{1}{\text{cos}^2 \text{ 60°}} - \text{tan}^2 \text{ 60°}\\[1em] = \dfrac{1}{\Big(\dfrac{1}{2}\Big)^2} - (\sqrt3)^2\\[1em] = \dfrac{4}{1} - 3\\[1em] = 4 - 3\\[1em] = 1$

Hence, $\dfrac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°} = 1$ .

Question 6

If 4 sin² θ - 1 = 0 and angle θ is less than 90°, find the value of θ and hence the value of cos² θ + tan² θ.

Answer

4 sin² θ - 1 = 0

⇒ 4 sin² θ = 1

⇒ sin² θ = $\dfrac{1}{4}$

⇒ sin θ = $\sqrt{\dfrac{1}{4}}$

⇒ sin θ = $\dfrac{1}{2}$

⇒ sin θ = sin 30°

So, θ = 30°

Now, cos² θ + tan² θ

= cos² 30° + tan² 30°

$= \Big(\dfrac{\sqrt3}{2}\Big)^2 + \Big(\dfrac{1}{\sqrt3}\Big)^2\\[1em] = \dfrac{3}{4} + \dfrac{1}{3}\\[1em] = \dfrac{3 \times 3}{4 \times 3} + \dfrac{1 \times 4}{3 \times 4}\\[1em] = \dfrac{9}{12} + \dfrac{4}{12}\\[1em] = \dfrac{9 + 4}{12}\\[1em] = \dfrac{13}{12}\\[1em] = 1\dfrac{1}{12}$

Hence, θ = 30° and cos² 30° + tan² 30° = $\dfrac{13}{12}$ = $1\dfrac{1}{12}$ .

Question 7

If sin 3A = 1 and 0 ≤ A ≤ 90°, find :

(i) sin A

(ii) cos 2 A

(iii) tan² A - $\dfrac{1}{\text{cos}^2 \text{ A}}$

Answer

sin 3A = 1

⇒ sin 3A = sin 90°

So, 3A = 90°

⇒ A = $\dfrac{90°}{3} = 30°$

(i) sin A = sin 30° = $\dfrac{1}{2}$

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

(ii) cos 2 A

= cos (2 x 30°)

= cos 60°

= $\dfrac{1}{2}$

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

(iii) tan² A - $\dfrac{1}{\text{cos}^2 \text{ A}}$

$= \text{tan}^2 \text{30°} - \dfrac{1}{\text{cos}^2 \text{30°}}\\[1em] = \Big(\dfrac{1}{\sqrt3}\Big)^2 - \dfrac{1}{\Big(\dfrac{\sqrt3}{2}\Big)^2}\\[1em] = \dfrac{1}{3} - \dfrac{1}{\dfrac{3}{4}}\\[1em] = \dfrac{1}{3} - \dfrac{4}{3}\\[1em] = \dfrac{1 - 4}{3}\\[1em] = \dfrac{- 3}{3}\\[1em] = - 1$

Hence, tan² A - $\dfrac{1}{\text{cos}^2 \text{ A}}$ = -1.

Question 8

If 2 cos 2A = ${\sqrt3}$ and A is acute, find :

(i) A

(ii) sin 3A

(iii) sin² (75° - A) + cos² (45° + A)

Answer

(i) 2 cos 2A = ${\sqrt3}$

⇒ cos 2A = $\dfrac{\sqrt3}{2}$

⇒ cos 2A = cos 30°

So, 2A = 30°

⇒ A = $\dfrac{30°}{2} = 15°$

Hence, A = 15°.

(ii) sin 3A

= sin (3 x 15°)

= sin 45°

= $\dfrac{1}{\sqrt2}$

Hence, sin 3A = $\dfrac{1}{\sqrt2}$ .

(iii) sin² (75° - A) + cos² (45° + A)

= sin² (75° - 15°) + cos² (45° + 15°)

= sin² 60° + cos² 60°

$= \Big(\dfrac{\sqrt3}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2\\[1em] = \dfrac{3}{4} + \dfrac{1}{4}\\[1em] = \dfrac{3 + 1}{4} \\[1em] = \dfrac{4}{4} \\[1em] = 1$

Hence, sin² (75° - A) + cos² (45° + A) = 1.

Exercise 21(F)

Question 1(a)

If sin A = cos A, the measurement of angle A is :

0°
30°
45°
60°

Answer

Given:

sin A = cos A

⇒ sin A = sin (90° - A)

So, A = 90° - A

⇒ A + A = 90°

⇒ 2A = 90°

⇒ A = $\dfrac{90°}{2}$

⇒ A = 45°

Hence, option 3 is the correct option.

Question 1(b)

If sin A = cos B and A ≠ B then the relation between angles A and B is :

A + B = 180°
A - B = 90°
B - A = 90°
A + B = 90°

Answer

Given:

sin A = cos B

⇒ sin A = sin (90° - B)

So, A = 90° - B

⇒ A + B = 90°

Hence, option 4 is the correct option.

Question 1(c)

If A + B = 90°, the value of

$\dfrac{\text{cos A}}{\text{sin B}}\times \dfrac{\text{tan B}}{\text{cot A}}$ is :

1
2
sin A
cos B

Answer

Given:

$\dfrac{\text{cos A}}{\text{sin B}}\times \dfrac{\text{tan B}}{\text{cot A}}\\[1em] ⇒ \dfrac{\text{cos A}}{\text{sin (90° - A)}}\times \dfrac{\text{tan (90° - A)}}{\text{cot A}}\\[1em] ⇒ \dfrac{\text{cos A}}{\text{cos A}}\times \dfrac{\text{cot A}}{\text{cot A}}\\[1em] ⇒ \dfrac{\cancel{cos A}}{\cancel{cos A}}\times \dfrac{\cancel{cot A}}{\cancel{cot A}}\\[1em] ⇒ 1$

Hence, option 1 is the correct option.

Question 1(d)

The value of :

cosec 40° cos 50° + sin 50° sec 40° is:

Answer

Given:

cosec 40° cos 50° + sin 50° sec 40°

= cosec 40° cos (90° - 40°) + sin (90° - 40°) sec 40°

= cosec 40° sin 40° + cos 40° sec 40°

= $\dfrac{1}{\text{sin 40°}} \times \text{sin 40°} + \text{cos 40°} \times \dfrac{1}{\text{cos 40°}}$

= 1 + 1

= 2

Hence, option 2 is the correct option.

Question 1(e)

In a triangle ABC, $\text{sec}\dfrac{A + C}{2}$ is equal to:

0
$\text{sec}\dfrac{B}{2}$
cosec B
$\text{cosec}\dfrac{B}{2}$

Answer

Given:

In Δ ABC,

$⇒ ∠ A + ∠ B + ∠ C = 180°\\[1em] ⇒ ∠ A + ∠ C = 180° - ∠ B\\[1em] ⇒ \dfrac{A + C}{2} = \dfrac{180° - B}{2}\\[1em] ⇒ \dfrac{A + C}{2} = 90° - \dfrac{B}{2}\\[1em] ⇒ \text{sec}\dfrac{A + C}{2} = \text{sec}{\Big(90° - \dfrac{B}{2}\Big)}\\[1em] ⇒ \text{sec}\dfrac{A + C}{2} = \text{cosec}\dfrac{B}{2}\\[1em]$

Hence, option 4 is the correct option.

Question 2(i)

Evaluate:

$\dfrac{\text{cos 22°}}{\text{sin 68°}}$

Answer

$\dfrac{\text{cos 22°}}{\text{sin 68°}} = \dfrac{\text{cos (90° - 68°)}}{\text{sin 68°}}\\[1em] = \dfrac{\text{sin 68°}}{\text{sin 68°}}\\[1em] = \dfrac{\cancel{sin 68°}}{\cancel{sin 68°}}\\[1em] = 1$

Hence, $\dfrac{\text{cos 22°}}{\text{sin 68°}}$ = 1.

Question 2(ii)

Evaluate:

$\dfrac{\text{tan 47°}}{\text{cot 43°}}$

Answer

$\dfrac{\text{tan 47°}}{\text{cot 43°}} = \dfrac{\text{tan (90° - 43°)}}{\text{cot 43°}}\\[1em] = \dfrac{\text{cot 43°}}{\text{cot 43°}}\\[1em] = \dfrac{\cancel{cot 43°}}{\cancel{cot 43°}}\\[1em] = 1$

Hence, $\dfrac{\text{tan 47°}}{\text{cot 43°}}$ = 1.

Question 2(iii)

Evaluate:

$\dfrac{\text{sec 75°}}{\text{cosec 15°}}$

Answer

Evaluate :

sin (90° - A) sin A - cos (90° - A) cos A

Answer

sin (90° - A) sin A - cos (90° - A) cos A

= cos A sin A - sin A cos A

= 0

Hence, sin (90° - A) sin A - cos (90° - A) cos A = 0.

Question 3(ii)

Evaluate :

sin² 35° - cos² 55°

Answer

sin² 35° - cos² 55°

= sin² (90° - 55°) - cos² 55°

= cos² 55° - cos² 55°

= 0

Hence, sin² 35° - cos² 55° = 0.

Question 3(iii)

Evaluate :

$\dfrac{\text{cot 54°}}{\text{tan 36°}}$ + $\dfrac{\text{tan 20°}}{\text{cot 70°}}$ - 2

Answer

$\dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2\\[1em] = \dfrac{\text{cot (90° - 36°)}}{\text{tan 36°}} + \dfrac{\text{tan (90° - 70°)}}{\text{cot 70°}} - 2\\[1em] = \dfrac{\text{tan 36°}}{\text{tan 36°}} + \dfrac{\text{cot 70°}}{\text{cot 70°}} - 2\\[1em] = \dfrac{\cancel{tan 36°}}{\cancel{tan 36°}} + \dfrac{\cancel{cot 70°}}{\cancel{cot 70°}} - 2\\[1em] = 1 + 1 - 2\\[1em] = 2 - 2\\[1em] = 0$

Hence, $\dfrac{\text{cot 54°}}{\text{tan 36°}} + \dfrac{\text{tan 20°}}{\text{cot 70°}} - 2 = 0$ .

Question 3(iv)

Evaluate :

cos² 25° - sin² 65° - tan² 45°

Answer

cos² 25° - sin² 65° - tan² 45°

= cos² (90° - 65°) - sin² 65° - tan² 45°

= sin² 65° - sin² 65° - tan² 45°

= - tan² 45°

= - 1

Hence, cos² 25° - sin² 65° - tan² 45° = -1.

Question 3(v)

Evaluate :

$\Big(\dfrac{\text{sin 77°}}{\text{cos 13°}}\Big)^2$ + $\Big(\dfrac{\text{cos 77°}}{\text{sin 13°}}\Big)^2$ - 2 cos² 45°

Answer

$\Big(\dfrac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{cos 77°}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\dfrac{\text{sin (90° - 13°)}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{cos (90° - 13°)}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\dfrac{\text{cos 13°}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{sin 13°}}{\text{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = \Big(\dfrac{\cancel{cos 13°}}{\cancel{cos 13°}}\Big)^2 + \Big(\dfrac{\cancel{sin 13°}}{\cancel{sin 13°}}\Big)^2 - \text{2 cos}^2 45°\\[1em] = 1^2 + 1^2 - 2 \times \Big(\dfrac{1}{\sqrt2}\Big)^2\\[1em] = 1 + 1 - 2 \times \Big(\dfrac{1}{2}\Big)\\[1em] = 2 - 1\\[1em] = 1$

Hence, $\Big(\dfrac{\text{sin 77°}}{\text{cos 13°}}\Big)^2 + \Big(\dfrac{\text{cos 77°}}{\text{sin 13°}}\Big) - \text{2 cos}^2 45° = 1$ .

Question 4(i)

Show that :

tan 10° tan 15° tan 75° tan 80° = 1

Answer

tan 10° tan 15° tan 75° tan 80° = 1

L.H.S. = tan 10° tan 15° tan 75° tan 80°

= tan (90° - 80°) tan (90° - 75°) tan 75° tan 80°

= cot 80° cot 75° tan 75° tan 80°

= $\dfrac{1}{\text{tan 80°}} \times \dfrac{1}{\text{tan 75°}} \times \text{tan 75°} \times \text{tan 80°}$

= $\dfrac{1}{\cancel{tan 80°}} \times \dfrac{1}{\cancel{tan 75°}} \times \cancel{tan 75°} \times \cancel{tan 80°}$

= 1

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence, tan 10° tan 15° tan 75° tan 80° = 1.

Question 4(ii)

Show that :

sin 42° sec 48° + cos 42° cosec 48° = 2

Answer

sin 42° sec 48° + cos 42° cosec 48° = 2

L.H.S. = sin 42° sec 48° + cos 42° cosec 48°

= sin (90° - 48°) sec 48° + cos (90° - 48°) cosec 48°

= cos 48° sec 48° + sin 48° cosec 48°

= $\text{cos 48°} \dfrac{1}{\text{cos 48°}} + \text{sin 48°} \dfrac{1}{\text{sin 48°}}$

= $\cancel{cos 48°} \dfrac{1}{\cancel{cos 48°}} + \cancel{sin 48°} \dfrac{1}{\cancel{sin 48°}}$

= 1 + 1

= 2

R.H.S. = 2

∴ L.H.S. = R.H.S.

Hence, sin 42° sec 48° + cos 42° cosec 48° = 2.

Question 5(i)

Express the following in terms of angles between 0° and 45° :

sin 59° + tan 63°

Answer

sin 59° + tan 63°

= sin (90° - 31°) + tan (90° - 27°)

= cos 31° + cot 27°

Hence, sin 59° + tan 63° = cos 31° + cot 27°.

Question 5(ii)

Express the following in terms of angles between 0° and 45° :

cosec 68° + cot 72°

Answer

cosec 68°+ cot 72°

= cosec (90° - 22°) + cot (90° - 18°)

= sec 22° + tan 18°

Hence, cosec 68°+ cot 72° = sec 22° + tan 18°.

Question 5(iii)

Express the following in terms of angles between 0° and 45° :

cos 74° + sec 67°

Answer

cos 74° + sec 67°

= cos (90° - 16°) + sec (90° - 23°)

= sin 16° + cosec 23°

Hence, cos 74° + sec 67° = sin 16° + cosec 23°.

Question 6

For triangle ABC, show that :

(i) $\text{sin}\dfrac{A+B}{2} = \text{cos}\dfrac{C}{2}$

(ii) $\text{tan}\dfrac{B+C}{2} = \text{cot}\dfrac{A}{2}$

Answer

(i) $\text{sin}\dfrac{A+B}{2} = \text{cos}\dfrac{C}{2}$

According to angle sum property,

$∠ A + ∠ B + ∠ C = 180°\\[1em] ∠ A + ∠ B = 180° - ∠ C\\[1em] ⇒ \dfrac{A + B}{2} = \dfrac{180° - C}{2}\\[1em]$

$\text{L.H.S.} = \text{sin}\dfrac{A+B}{2}\\[1em] = \text{sin}\dfrac{180° - C}{2}\\[1em] = \text{sin}\Big(90° - \dfrac{C}{2}\Big)\\[1em] = \text{cos}\dfrac{C}{2}$

R.H.S. = $\text{cos}\dfrac{C}{2}$

∴ L.H.S. = R.H.S.

Hence, $\text{sin}\dfrac{A+B}{2} = \text{cos}\dfrac{C}{2}$ .

(ii) $\text{tan}\dfrac{B+C}{2} = \text{cot}\dfrac{A}{2}$

According to angle sum property,

$∠ A + ∠ B + ∠ C = 180°\\[1em] ∠ B + ∠ C = 180° - ∠ A\\[1em] ⇒ \dfrac{B + C}{2} = \dfrac{180° - A}{2}\\[1em]$

$\text{L.H.S.} = \text{tan}\dfrac{B+C}{2}\\[1em] = \text{tan}\Big(\dfrac{180° - A}{2}\Big)\\[1em] = \text{tan}\Big(90° - \dfrac{A}{2}\Big)\\[1em] = \text{cot}\dfrac{A}{2}$

R.H.S. = $\text{cot}\dfrac{A}{2}$

∴ L.H.S. = R.H.S.

Hence, $\text{tan}\dfrac{B+C}{2} = \text{cot}\dfrac{A}{2}$ .

Question 7(i)

Evaluate :

$\text{3}\dfrac{\text{ sin 72°}}{\text{ cos 18°}}$ - $\dfrac{\text{sec 32°}}{\text{cosec 58°}}$

Answer

$\text{3}\dfrac{\text{ sin 72°}}{\text{ cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}}\\[1em] = \text{3}\dfrac{\text{ sin (90° - 18°)}}{\text{ cos 18°}} - \dfrac{\text{sec (90° - 58°)}}{\text{cosec 58°}}\\[1em] = \text{3}\dfrac{\text{ cos 18°}}{\text{ cos 18°}} - \dfrac{\text{cosec 58°}}{\text{cosec 58°}}\\[1em] = \text{3}\dfrac{\cancel{ cos 18°}}{\cancel{ cos 18°}} - \dfrac{\cancel{cosec 58°}}{\cancel{cosec 58°}}\\[1em] = 3 - 1\\[1em] = 2$

Hence, $\text{3}\dfrac{\text{ sin 72°}}{\text{ cos 18°}} - \dfrac{\text{sec 32°}}{\text{cosec 58°}} = 2$ .

Question 7(ii)

Evaluate :

3 cos 80° cosec 10° + 2 sin 59° sec 31°

Answer

3 cos 80° cosec 10° + 2 sin 59° sec 31°

= 3 cos (90° - 10°) cosec 10° + 2 sin (90° - 31°) sec 31°

= 3 sin 10° cosec 10° + 2 cos 31° sec 31°

= $3 \text{sin 10°} \times \dfrac{1}{\text{sin 10°}} + 2 \text{cos 31°} \times \dfrac{1}{\text{cos 31°}}$

= $3 \cancel{sin 10°} \times \dfrac{1}{\cancel{sin 10°}} + 2 \cancel{cos 31°} \times \dfrac{1}{\cancel{cos 31°}}$

Evaluate :

$2\dfrac{\text{tan 57°}}{\text{cot 33°}}$ - $\dfrac{\text{cot 70°}}{\text{tan 20°}}$ - ${\sqrt2} \text{cos 45°}$

Answer

$2\dfrac{\text{tan 57°}}{\text{cot 33°}} - \dfrac{\text{cot 70°}}{\text{tan 20°}} - {\sqrt2} \text{cos45°}\\[1em] = 2\dfrac{\text{tan (90° - 33°)}}{\text{cot 33°}} - \dfrac{\text{cot (90° - 20°)}}{\text{tan 20°}} - {\sqrt2} \text{cos45°}\\[1em] = 2\dfrac{\text{cot 33°}}{\text{cot 33°}} - \dfrac{\text{tan 20°}}{\text{tan 20°}} - {\sqrt2} \dfrac{1}{\sqrt2}\\[1em] = 2\dfrac{\cancel{cot 33°}}{\cancel{cot 33°}} - \dfrac{\cancel{tan 20°}}{\cancel{tan 20°}} - \cancel{\sqrt2} \dfrac{1}{\cancel{\sqrt2}}\\[1em] = 2 - 1 - 1\\[1em] = 0$

Hence, $2\dfrac{\text{tan 57°}}{\text{cot 33°}} - \dfrac{\text{cot 70°}}{\text{tan 20°}} - {\sqrt2} \text{cos45°} = 0$ .

Question 7(vii)

Evaluate :

$\dfrac{\text{cot}^2 \text{ 41°}}{\text{tan}^2 \text{ 49°}}$ - $2\dfrac{\text{ sin}^2 \text{ 75°}}{\text{ cos}^2 \text{ 15°}}$

Answer

$\dfrac{\text{cot}^2 \text{ 41°}}{\text{tan}^2 \text{ 49°}} - 2\dfrac{\text{ sin}^2 \text{ 75°}}{\text{ cos}^2 \text{ 15°}}\\[1em] = \dfrac{\text{cot}^2 \text{ (90° - 49°)}}{\text{tan}^2 \text{ 49°}} - 2\dfrac{\text{ sin}^2 \text{ (90° - 15°)}}{\text{ cos}^2 \text{ 15°}}\\[1em] = \dfrac{\text{tan}^2 \text{49°}}{\text{tan}^2 \text{ 49°}} - 2\dfrac{\text{ cos}^2 \text{15°}}{\text{ cos}^2 \text{ 15°}}\\[1em] = \dfrac{\cancel{tan^2 49°}}{\cancel{tan^2 49°}} - 2\dfrac{\cancel{ cos^2 15°}}{\cancel{ cos^2 15°}}\\[1em] = 1 - 2\\[1em] = - 1$

Hence, $\dfrac{\text{cot}^2 \text{ 41°}}{\text{tan}^2 \text{ 49°}} - 2\dfrac{\text{ sin}^2 \text{ 75°}}{\text{ cos}^2 \text{ 15°}} = -1$ .

Question 7(viii)

Evaluate :

$\dfrac{\text{cos 70°}}{\text{sin 20°}}$ + $\dfrac{\text{cos 59°}}{\text{sin 31°}}$ - 8 sin² 30°

Answer

$\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 \text{30°}\\[1em] =\dfrac{\text{cos (90° - 70°)}}{\text{sin 20°}} + \dfrac{\text{cos (90° - 31°)}}{\text{sin 31°}} - 8 \times \Big(\dfrac{1}{2}\Big)^2\\[1em] =\dfrac{\text{sin 70°}}{\text{sin 20°}} + \dfrac{\text{sin 31°}}{\text{sin 31°}} - 8 \times \Big(\dfrac{1}{4}\Big)\\[1em] =\dfrac{\cancel{sin 70°}}{\cancel{sin 20°}} + \dfrac{\cancel{sin 31°}}{\cancel{sin 31°}} - 2\\[1em] = 1 + 1 - 2\\[1em] = 0$

Hence, $\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 \text{30°} = 0$ .

Question 7(ix)

Evaluate :

14 sin 30° + 6 cos 60° - 5 tan 45°.

Answer

14 sin 30° + 6 cos 60° - 5 tan 45°

$= 14 \times \dfrac{1}{2} + 6 \times \dfrac{1}{2} - 5 \times 1$

= 7 + 3 - 5

= 10 - 5

= 5

Hence, 14 sin 30° + 6 cos 60° - 5 tan 45° = 5.

Question 8

A triangle ABC is right-angled at B; find the value of $\dfrac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}}$ .

Answer

Given:

ABC is right-angled triangle at B.

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ A + 90° + ∠ C = 180°

⇒ ∠ A + ∠ C = 180° - 90°

⇒ ∠ A + ∠ C = 90°

⇒ ∠ A = 90° - ∠ C

Now,

$\dfrac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}}\\[1em] = \dfrac{\text{sec (90° - C) . sin C - tan (90° - C). tan C}}{\text{sin B}}\\[1em] = \dfrac{\text{cosec C . sin C - cot C. tan C}}{\text{sin B}}\\[1em] = \dfrac{\dfrac{1}{\text{sin C}} \times \text{sin C} - \dfrac{1}{\text{tan C}} \times \text{tan C}}{\text{sin B}}\\[1em] = \dfrac{\dfrac{1}{\cancel{sin C}} \times \cancel{sin C} - \dfrac{1}{\cancel{tan C}} \times \cancel{tan C}}{\text{sin B}}\\[1em] = \dfrac{1 - 1}{\text{sin B}}\\[1em] = 0$

Hence, $\dfrac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}} = 0$ .

Question 9

In each case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.

(i) sin (90° - 3A) . cosec 42° = 1

(ii) cos (90° - A) . sec 77° = 1

Answer

(i) sin (90° - 3A) . cosec 42° = 1

⇒ cos 3A . cosec 42° = 1

⇒ cos 3A x $\dfrac{1}{\text{sin 42°}}$ = 1

⇒ cos 3A = sin 42°

⇒ cos 3A = sin (90° - 48°)

⇒ cos 3A = cos 48°

So, 3A = 48°

⇒ A = $\dfrac{48°}{3}$

⇒ A = 16°

Hence, A = 16°.

(ii) cos (90° - A) . sec 77° = 1

⇒ sin A . sec 77° = 1

⇒ sin A x $\dfrac{1}{\text{cos 77°}}$ = 1

⇒ sin A = cos 77°

⇒ sin A = cos (90° - 13°)

⇒ sin A = sin 13°

So, A = 13°

Hence, A = 13°.

Test Yourself

Question 1(a)

The value of tan A is :

$\dfrac{5}{12}$
$\dfrac{12}{13}$
$\dfrac{5}{13}$
$\dfrac{13}{12}$

Answer

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

⇒ Hypotenuse² = Base² + Height²

⇒ PQ² = QR² + PR²

⇒ 13² = PR² + 12²

⇒ 169 = PR² + 144

⇒ PR² = 169 - 144

⇒ PR² = 25

⇒ PR = $\sqrt{25}$

⇒ PR = 5 cm.

By formula, tan A = $\dfrac{\text{Height}}{\text{Base}}$

From figure,

$\text{tan A} = \dfrac{PR}{RQ} \\[1em] = \dfrac{5}{12}.$

Hence, option 1 is the correct option.

Question 1(b)

The value of sin B - cos A is;

$\dfrac{1}{2}$
1
0
none of these

Answer

Since ΔABC is a right angled triangle.

By formula,

sin θ = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$ and cos θ = $\dfrac{\text{Base}}{\text{Hypotenuse}}$ .

So, sin B = $\dfrac{\text{AC}}{\text{AB}} = \dfrac{3}{5}$

cos A = $\dfrac{\text{AC}}{\text{AB}} = \dfrac{3}{5}$

The value of sin B - cos A = $\dfrac{3}{5} - \dfrac{3}{5}$ = 0.

Hence, option 3 is the correct option.

Question 1(c)

The value of cos 60° - sin 90° + 2 cos 0° is :

$\dfrac{1}{2}$
$-\dfrac{1}{2}$
$-1\dfrac{1}{2}$
$1\dfrac{1}{2}$

Answer

Solving,

$\Rightarrow \text{cos 60° - sin 90° + 2 cos 0° } = \dfrac{1}{2} - 1 + 2 \times 1\\[1em] = \dfrac{1}{2} - 1 + 2\\[1em] = \dfrac{1}{2} + 1\\[1em] = \dfrac{1 + 2}{2}\\[1em] = \dfrac{3}{2}\\[1em] = 1\dfrac{1}{2}.$

Hence, option 4 is the correct option.

Question 1(d)

The value of sin 23° - cos 67° is :

1
0
cos 44°
-cos 44°

Answer

Given,

⇒ sin 23° - cos 67°

⇒ sin 23° - cos (90° - 23°)

⇒ sin 23° - sin 23°

⇒ 0.

Hence, option 2 is the correct option.

Question 1(e)

Statement 1: The angle C of a right angled triangle is 90°, then tan A = cot B.

Statement 2: Since, angle C of triangle ABC = 90°.

∴ ∠A + ∠B = 90° ⇒ ∠A = 90° - ∠B

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given, ∠C = 90°.

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180°

⇒ ∠A + ∠B = 180° - 90°

⇒ ∠A + ∠B = 90°

⇒ ∠A = 90° - ∠B

So, statement 2 is true.

⇒ tan A = tan (90° - ∠B)

⇒ tan A = cot B

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Question 1(f)

Statement 1: If 3 cos A = 4, sec A = $\dfrac{3}{4}$ .

Statement 2: 3 cos A = 4 ⇒ cos A = $\dfrac{4}{3}$ and sec A = $\dfrac{1}{\text{cos A}} = \dfrac{3}{4}$ .

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
Statement 1 is false, and statement 2 is true.

Answer

Given,

⇒ 3 cos A = 4

⇒ cos A = $\dfrac{4}{3}$

⇒ $\dfrac{1}{\text{cos A}} = \dfrac{1}{\dfrac{4}{3}}$

⇒ sec A = $\dfrac{3}{4}$ .

∴ Both the statements are true.

Hence, option 1 is the correct option.

Question 1(g)

Assertion (A): The value of sin² 30° - 2 cos³ 60° + 2 tan⁴ 45° is 2.

Reason (R): sin 30° = $\dfrac{1}{2}$ , cos 60° = $\dfrac{1}{2}$ , and tan 45° = 1

A is true, but R is false.
A is false, but R is true.
Both A and R are true, and R is the correct reason for A.
Both A and R are true, and R is the incorrect reason for A.

Answer

We know that,

sin 30° = $\dfrac{1}{2}$ , cos 60° = $\dfrac{1}{2}$ , and tan 45° = 1.

So, reason (R) is true.

$\text{sin}^2 30° - \text{2 cos}^3 60° + \text{2 tan}^4 45° = \Big(\dfrac{1}{2}\Big)^2 - 2 \times \Big(\dfrac{1}{2}\Big)^3 + 2 \times 1^4\\[1em] = \dfrac{1}{4} - 2 \times \dfrac{1}{8} + 2\\[1em] = \dfrac{1}{4} - \dfrac{1}{4} + 2\\[1em] = 2.$

So, assertion (A) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Question 1(h)

Assertion (A): If A = 30°, the value of 4 sin A sin (60° - A) sin (60° + A) = 1.

Reason (R): 60° - A = 30° and 60° + A = 90°.

A is true, but R is false.
A is false, but R is true.
Both A and R are true, and R is the correct reason for A.
Both A and R are true, and R is the incorrect reason for A.

Answer

If A = 30°,

60° - A = 60° - 30° = 30° and 60° + A = 60° + 30° = 90°.

So, reason (R) is true.

sin A = sin 30° = $\dfrac{1}{2}$

sin (60° - A) = sin 30° = $\dfrac{1}{2}$

sin (60° + A) = sin 90° = 1

Substituting values in 4 sin A sin (60° - A) sin (60° + A), we get :

$\Rightarrow 4 \times \dfrac{1}{2} \times \dfrac{1}{2} \times 1\\[1em] \Rightarrow 4 \times \dfrac{1}{4}\\[1em] \Rightarrow 1.$

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

Question 2

If $\text{cosec θ} = {\sqrt5}$ find the value of :

(i) 2 - sin² θ - cos² θ

(ii) $2 + \dfrac{1}{\text{sin}^2 \text{ θ}} - \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}$

Answer

Given:

$\text{cosec θ} = {\sqrt5}$

$⇒ \text{cosec θ} = \dfrac{Hypotenuse}{Perpendicular} = {\sqrt5}\\[1em]$

If cosec θ = 5 find the value of : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

∴ If length of AC = $\sqrt{5}$ x unit, length of BC = x unit.

In Δ ABC,

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

⇒ ( $\sqrt{5}$ x)² = (x)² + AB²

⇒ 5x² = x² + AB²

⇒ AB² = 5x² - x²

⇒ AB² = 4x²

⇒ AB = $\sqrt{4 \text{x}^2}$

⇒ AB = 2x

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

$= \dfrac{BC}{CA} = \dfrac{x}{\sqrt{5}x} = \dfrac{1}{\sqrt{5}}$

cos θ = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC} = \dfrac{2x}{\sqrt{5}x} = \dfrac{2}{\sqrt{5}}$

Now,

2 - sin²θ - cos²θ

$= 2 - \Big(\dfrac{1}{\sqrt{5}}\Big)^2 - \Big(\dfrac{2}{\sqrt{5}}\Big)^2\\[1em] = 2 - \dfrac{1}{5} - \dfrac{4}{5}\\[1em] = 2 + \dfrac{-1 - 4}{5}\\[1em] = 2 + \dfrac{-5}{5}\\[1em] = 2 - 1\\[1em] = 1$

Hence, 2 - sin²θ - cos²θ = 1.

(ii) $2 + \dfrac{1}{\text{sin}^2 \text{ θ}} - \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}$

$= 2 + \dfrac{1}{\Big(\dfrac{1}{\sqrt{5}}\Big)^2} - \dfrac{\Big(\dfrac{2}{\sqrt{5}}\Big)^2}{\Big(\dfrac{1}{\sqrt{5}}\Big)^2}\\[1em] = 2 + \dfrac{1}{\dfrac{1}{5}} - \dfrac{\dfrac{4}{5}}{\dfrac{1}{5}}\\[1em] = 2 + \dfrac{5}{1} - \dfrac{\dfrac{4}{\cancel{5}}}{\dfrac{1}{\cancel{5}}}\\[1em] = 2 + 5 - \dfrac{4}{1}\\[1em] = 7 - 4\\[1em] = 3$

Hence, $2 + \dfrac{1}{\text{sin}^2 \text{ θ}} - \dfrac{\text{cos}^2 \text{ θ}}{\text{sin}^2 \text{ θ}}$ = 3.

Question 3

In the given figure; ∠C = 90° and D is mid-point of AC. Find :

(i) $\dfrac{\text{tan ∠CAB}}{\text{tan ∠CDB}}$

(ii) $\dfrac{\text {tan ∠ABC}}{\text {tan ∠DBC}}$

In the given figure; ∠C = 90° and D is mid-point of AC. Find : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

Answer

Since D is the mid-point of A. So, AC = 2DC

(i) $\text{tan ∠CAB} = \dfrac{Perpendicular}{Base} = \dfrac{BC}{AC}$

$\text{tan ∠CDB} = \dfrac{Perpendicular}{Base} = \dfrac{BC}{DC}$

Now,

$\dfrac{\text{tan ∠CAB}}{\text{tan ∠CDB}}\\[1em] = \dfrac{\dfrac{BC}{AC}}{\dfrac{BC}{DC}}\\[1em] = \dfrac{BC \times DC}{AC \times BC}\\[1em] = \dfrac{\cancel{BC} \times DC}{AC \times \cancel{BC}}\\[1em] = \dfrac{DC}{AC}\\[1em] = \dfrac{DC}{2 \times DC}\\[1em] = \dfrac{\cancel{DC}}{2 \times \cancel{DC}}\\[1em] = \dfrac{1}{2}$

Hence, $\dfrac{\text{tan ∠CAB}}{\text{tan ∠CDB}} = \dfrac{1}{2}$

(ii) $\text{tan ∠ABC} = \dfrac{Perpendicular}{Base} = \dfrac{AC}{BC}$

$\text{tan ∠DBC} = \dfrac{Perpendicular}{Base} = \dfrac{DC}{BC}$

Now,

$\dfrac{\text {tan ∠ABC}}{\text {tan ∠DBC}}\\[1em] = \dfrac{\dfrac{AC}{BC}}{\dfrac{DC}{BC}}\\[1em] = \dfrac{AC \times BC}{BC \times DC}\\[1em] = \dfrac{AC \times \cancel{BC}}{\cancel{BC} \times DC}\\[1em] = \dfrac{AC}{DC}\\[1em] = \dfrac{2 \times DC}{DC}\\[1em] = \dfrac{2 \times \cancel{DC}}{\cancel{DC}}\\[1em] = 2$

Hence, $\dfrac{\text {tan ∠ABC}}{\text {tan ∠DBC}} = 2$

Question 4

If 3 cos A = 4 sin A, find the value of :

(i) cos A

(ii) 3 - cot² A + cosec² A

Answer

Given:

3 cos A = 4 sin A

$\dfrac{\text{cos A}}{\text{sin A}} = \dfrac{4}{3}$

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

$\text{cot A} = \dfrac{Base}{Perpendicular} = \dfrac{4}{3}$

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

If 3 cos A = 4 sin A, find the value of : Trigonometrical Ratios, Concise Mathematics Solutions ICSE Class 9.

In Δ ABC,

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

⇒ AC² = (4x)² + (3x)²

⇒ AC² = 16x² + 9x²

⇒ AC² = 25x²

⇒ AC = $\sqrt{25\text{x}^2}$

⇒ AC = 5x

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

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

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

(ii) 3 - cot² A + cosec² A

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

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

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

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

Now,

3 - cot² A + cosec² A

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

Hence, 3 - cot² A + cosec² A = 4.

Question 5

Use the information given in the following figure to evaluate :

$\dfrac{10}{\text{sin x}} + \dfrac{6}{\text{sin y}} - \text{6 cot y}$ .

Answer

From the figure, in Δ ADC,

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

⇒ 20² = DC² + 12²

⇒ 400 = DC² + 144

⇒ DC² = 400 - 144

⇒ DC² = 256

⇒ DC = $\sqrt{256}$

⇒ DC = 16

BD = BC - DC

= 21 - 16 = 5

In Δ ABD,

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

⇒ AB² = 12² + 5²

⇒ AB² = 144 + 25

⇒ AB² = 169

⇒ AB = $\sqrt{169}$

⇒ AB = 13

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

$= \dfrac{BD}{AB} = \dfrac{5}{13}$

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

$= \dfrac{AD}{AC} = \dfrac{12}{20} = \dfrac{3}{5}$

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

$= \dfrac{DC}{AD} = \dfrac{16}{12} = \dfrac{4}{3}$

Now,

$\dfrac{10}{\text{sin x}} + \dfrac{6}{\text{sin y}} - \text{6 cot y}\\[1em] = \dfrac{10}{\dfrac{5}{13}} + \dfrac{6}{\dfrac{3}{5}} - 6 \times \dfrac{4}{3}\\[1em] = \dfrac{10 \times 13}{5} + \dfrac{6 \times 5}{3} - \dfrac{24}{3}\\[1em] = \dfrac{130}{5} + \dfrac{30}{3} - \dfrac{24}{3}\\[1em] = 26 + 10 - 8\\[1em] = 28$

Hence, $\dfrac{10}{\text{sin x}} + \dfrac{6}{\text{sin y}} - \text{6 cot y} = 28.$

Question 6(i)

Evaluate :

$\dfrac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}}$ , when A = 15°.

Answer

$\dfrac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}}\\[1em] = \dfrac{\text{cos (3 x 15°) - 2 cos (4 x 15°)}}{\text{sin (3 x 15°) + 2 sin (4 x 15°)}}\\[1em] = \dfrac{\text{cos 45° - 2 cos 60°}}{\text{sin 45° + 2 sin 60°}}\\[1em] = \dfrac{\Big(\dfrac{1}{\sqrt2}\Big) - 2 \times \Big(\dfrac{1}{2}\Big)}{\Big(\dfrac{1}{\sqrt2}\Big) + 2 \times \Big(\dfrac{\sqrt3}{2}\Big)}\\[1em] = \dfrac{\dfrac{1}{\sqrt2} - 1}{\dfrac{1}{\sqrt2} + \sqrt3}\\[1em] = \dfrac{\dfrac{1}{\sqrt2} - \dfrac{1 \times \sqrt2}{\sqrt2}}{\dfrac{1}{\sqrt2} + \dfrac{\sqrt3 \times \sqrt2}{\sqrt2}}\\[1em] = \dfrac{\dfrac{1}{\sqrt2} - \dfrac{\sqrt2}{\sqrt2}}{\dfrac{1}{\sqrt2} + \dfrac{\sqrt6}{\sqrt2}}\\[1em] = \dfrac{\dfrac{1 - \sqrt2}{\sqrt2}}{\dfrac{1 + \sqrt6}{\sqrt2}}\\[1em] = \dfrac{\dfrac{1 - \sqrt2}{\cancel{\sqrt2}}}{\dfrac{1 + \sqrt6}{\cancel{\sqrt2}}}\\[1em] = \dfrac{1 - \sqrt2}{1 + \sqrt6}\\[1em] = \dfrac{(1 - \sqrt2) \times (1 - \sqrt6)}{(1 + \sqrt6) \times (1 - \sqrt6)}\\[1em] = \dfrac{1 - \sqrt2 - \sqrt6 + \sqrt{12}}{1 - 6}\\[1em] = \dfrac{1 - \sqrt2 - \sqrt6 + 2\sqrt3}{-5}\\[1em] = \dfrac{1}{5}(- 1 + \sqrt2 + \sqrt6 - 2\sqrt3)$

Hence, $\dfrac{\text{cos 3 A - 2 cos 4 A}}{\text{sin 3 A + 2 sin 4 A}} = \dfrac{1}{5}(- 1 + \sqrt2 + \sqrt6 - 2\sqrt3)$

Question 6(ii)

Evaluate :

$\dfrac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}}$ ; when B = 20°.

Answer

$\dfrac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}}\\[1em] = \dfrac{\text{3 sin (3 x 20°) + 2 cos((2 x 20°) + 5°)}}{\text{2 cos (3 x 20°) - sin((2 x 20°) - 10°)}}\\[1em] = \dfrac{\text{3 sin 60° + 2 cos(40° + 5°)}}{\text{2 cos 60° - sin(40° - 10°)}}\\[1em] = \dfrac{\text{3 sin 60° + 2 cos 45°}}{\text{2 cos 60° - sin 30°}}\\[1em] = \dfrac{3 \times \dfrac{\sqrt3}{2} + 2 \times \dfrac{1}{\sqrt2}}{2 \times \dfrac{1}{2} - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3}{2} + \dfrac{2}{\sqrt2}}{\dfrac{2}{2} - \dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3}{2} + \dfrac{2 \times \sqrt2}{\sqrt2 \times \sqrt2}}{\dfrac{2 - 1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3}{2} + \dfrac{2\sqrt2}{2}}{\dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3 + 2\sqrt2}{2}}{\dfrac{1}{2}}\\[1em] = \dfrac{\dfrac{3\sqrt3 + 2\sqrt2}{\cancel2}}{\dfrac{1}{\cancel2}}\\[1em] = 3\sqrt3 + 2\sqrt2$

Hence, $\dfrac{\text{3 sin 3 B + 2 cos(2 B + 5°)}}{\text{2 cos 3 B - sin(2 B - 10°)}} = 3\sqrt3 + 2\sqrt2.$

Question 7(i)

Solve for x :

2 cos 3x - 1 = 0

Answer

2 cos 3x - 1 = 0

⇒ 2 cos 3x = 1

⇒ cos 3x = $\dfrac{1}{2}$

⇒ cos 3x = cos 60°

So, 3x = 60°

⇒ x = $\dfrac{60°}{3} = 20°$

Hence, x = 20°.

Question 7(ii)

Solve for x :

cos $\dfrac{x}{3} - 1 = 0$

Answer

$\text{cos }\dfrac{x}{3} - 1 = 0$

$⇒ \text{cos } \dfrac{x}{3} = 1\\[1em] ⇒ \text{cos } \dfrac{x}{3} = \text{cos 0°}$

So,

$⇒ \dfrac{x}{3} = 0°\\[1em] ⇒ x = 3 \times 0°\\[1em] ⇒ x = 0°\\[1em]$

Hence, x = 0°.

Question 7(iii)

Solve for x :

sin (x + 10°) = $\dfrac{1}{2}$

Answer

sin (x + 10°) = $\dfrac{1}{2}$

⇒ sin (x + 10°) = sin 30°

So, x + 10° = 30°

⇒ x = 30° - 10° = 20°

Hence, x = 20°.

Question 7(iv)

Solve for x :

cos (2x - 30°) = 0

Answer

cos (2x - 30°) = 0

⇒ cos (2x - 30°) = cos 90°

So, 2x - 30° = 90°

⇒ 2x = 90° + 30°

⇒ x = $\dfrac{120°}{2}$

⇒ x = 60°

Hence, x = 60°.

Question 7(v)

Solve for x :

2 cos (3x - 15°) = 1

Answer

2 cos (3x - 15°) = 1

⇒ cos (3x - 15°) = $\dfrac{1}{2}$

⇒ cos (3x - 15°) = cos 60°

So, 3x - 15° = 60°

⇒ 3x = 60° + 15°

⇒ x = $\dfrac{75°}{3}$

⇒ x = 25°

Hence, x = 25°.

Question 7(vi)

Solve for x :

tan² (x - 5°) = 3

Answer

tan² (x - 5°) = 3

⇒ tan (x - 5°) = $\sqrt3$

⇒ tan (x - 5°) = tan 60°

So, x - 5° = 60°

⇒ x = 60° + 5°

⇒ x = 65°

Hence, x = 65°.

Question 7(vii)

Solve for x :

3 tan² (2x - 20°) = 1

Answer

3 tan² (2x - 20°) = 1

⇒ tan² (2x - 20°) = $\dfrac{1}{3}$

⇒ tan (2x - 20°) = $\sqrt\dfrac{1}{3}$

⇒ tan (2x - 20°) = $\dfrac{1}{\sqrt3}$

⇒ tan (2x - 20°) = tan 30°

So, 2x - 20° = 30°

⇒ 2x = 30° + 20°

⇒ 2x = 50°

⇒ x = $\dfrac{50°}{2}$

⇒ x = 25°

Hence, x = 25°.

Question 7(viii)

Solve for x :

$cos\Big(\dfrac{x}{2} + 10°\Big)$ = $\dfrac{\sqrt3}{2}$

Answer

$\text{cos}\Big(\dfrac{x}{2} + 10°\Big)$ = $\dfrac{\sqrt3}{2}$

$⇒ \text{cos } \Big(\dfrac{x}{2} + 10°\Big) = \text{cos 30°}$

So,

$⇒ \Big(\dfrac{x}{2} + 10°\Big) = 30°\\[1em] ⇒ \dfrac{x}{2} = 30° - 10°\\[1em] ⇒ x = 20° \times 2\\[1em] ⇒ x = 40° \\[1em]$

Hence, x = 40°.

Question 7(ix)

Solve for x :

sin² x + sin² 30° = 1

Answer

sin² x + sin² 30° = 1

$⇒ \text{sin }^2 x + \Big(\dfrac{1}{2}\Big)^2 = 1\\[1em] ⇒ \text{sin }^2 x + \Big(\dfrac{1}{4}\Big) = 1\\[1em] ⇒ \text{sin }^2 x = 1 - \Big(\dfrac{1}{4}\Big)\\[1em] ⇒ \text{sin }^2 x = \dfrac{4}{4} - \dfrac{1}{4}\\[1em] ⇒ \text{sin }^2 x = \dfrac{4 - 1}{4}\\[1em] ⇒ \text{sin }^2 x = \dfrac{3}{4}\\[1em] ⇒ \text{sin } x = \sqrt\dfrac{3}{4}\\[1em] ⇒ \text{sin } x = \dfrac{\sqrt3}{2}\\[1em] ⇒ \text{sin } x = \text{sin } 60°\\[1em]$

So, x = 60°

Hence, x = 60°.

Question 7(x)

Solve for x :

cos² 30° + cos² x = 1

Answer

cos² 30° + cos² x = 1

$⇒ \Big(\dfrac{1}{2}\Big)^2 + \text{cos }^2 x = 1\\[1em] ⇒ \Big(\dfrac{1}{4}\Big) + \text{cos }^2 x = 1\\[1em] ⇒ \text{cos }^2 x = 1 - \Big(\dfrac{1}{4}\Big)\\[1em] ⇒ \text{cos }^2 x = \dfrac{4}{4} - \dfrac{1}{4}\\[1em] ⇒ \text{cos }^2 x = \dfrac{4 - 1}{4}\\[1em] ⇒ \text{cos } x = \sqrt\dfrac{3}{4}\\[1em] ⇒ \text{cos } x = \dfrac{\sqrt3}{2}\\[1em] ⇒ \text{cos } x = \text{cos 60°} \\[1em]$

So, x = 60°

Hence, x = 60°.

Question 7(xi)

Solve for x :

cos² 30° + sin² 2x = 1

Answer

cos² 30° + sin² 2x = 1

$⇒ \Big(\dfrac{\sqrt3}{2}\Big)^2 + \text{sin }^2 2x = 1\\[1em] ⇒ \Big(\dfrac{3}{4}\Big) + \text{sin }^2 2x = 1\\[1em] ⇒ \text{sin}^2 2x = 1 - \dfrac{3}{4}\\[1em] ⇒ \text{sin}^2 2x = \dfrac{4}{4} - \dfrac{3}{4}\\[1em] ⇒ \text{sin}^2 2x = \dfrac{4 - 3}{4}\\[1em] ⇒ \text{sin}^2 2x = \dfrac{1}{4}\\[1em] ⇒ \text{sin} 2x = \sqrt\dfrac{1}{4}\\[1em] ⇒ \text{sin} 2x = \dfrac{1}{2}\\[1em] ⇒ \text{sin} 2x = \text{sin }30°$

So, 2x = 30°

⇒ x = $\dfrac{30°}{2}$

⇒ x = 15°

Hence, x = 15°.

Question 7(xii)

Solve for x :

sin² 60° + cos² (3x - 9°) = 1

Answer

sin² 60° + cos² (3x - 9°) = 1

$⇒ \Big(\dfrac{\sqrt3}{2}\Big)^2 + \text{cos }^2 (3x - 9°) = 1\\[1em] ⇒ \Big(\dfrac{3}{4}\Big) + \text{cos }^2 (3x - 9°) = 1\\[1em] ⇒ \text{cos}^2 (3x - 9°) = 1 - \dfrac{3}{4}\\[1em] ⇒ \text{cos}^2 (3x - 9°) = \dfrac{4}{4} - \dfrac{3}{4}\\[1em] ⇒ \text{cos}^2 (3x - 9°) = \dfrac{4 - 3}{4} \\[1em] ⇒ \text{cos}^2 (3x - 9°) = \dfrac{1}{4} \\[1em] ⇒ \text{cos} (3x - 9°) = \sqrt\dfrac{1}{4} \\[1em] ⇒ \text{cos} (3x - 9°) = \dfrac{1}{2} \\[1em] ⇒ \text{cos} (3x - 9°) = \text{cos } 60°$

So, 3x - 9° = 60°

⇒ 3x = 60° + 9°

⇒ 3x = 69°

⇒ x = $\dfrac{69°}{3}$

⇒ x = 23°

Hence, x = 23°.

Question 8

If 2 cos (A + B) = 2 sin (A - B) = 1; find the values of A and B.

Answer

2 cos (A + B) = 1

⇒ cos (A + B) = $\dfrac{1}{2}$

⇒ cos (A + B) = cos 60°

So, A + B = 60° ...............(1)

2 sin (A - B) = 1

⇒ sin (A - B) = $\dfrac{1}{2}$

⇒ sin (A - B) = sin 30°

So, A - B = 30° ...............(2)

Adding equation (1) and (2), we get

(A + B) + (A - B) = 60° + 30°

⇒ A + B + A - B = 90°

⇒ 2A = 90°

⇒ A = $\dfrac{90°}{2}$

⇒ A = 45°

From equation (2), A - B = 30°

⇒ 45° - B = 30°

⇒ B = 45° - 30°

⇒ B = 15°

Hence, A = 45° and B = 15°.

Question 9

For the triangle ABC, show that

$\text{sin}^2\dfrac{A}{2}$ + $\text{sin}^2\dfrac{B+C}{2}$ = 1.

Answer

For triangle ABC,

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ B + ∠ C = 180° - ∠ A

⇒ $\dfrac{B + C}{2} = \dfrac{180° - A}{2}$

⇒ $\dfrac{B + C}{2} = 90° - \dfrac{A}{2}$

$\text{L.H.S.} = \text{sin}^2\dfrac{A}{2} + \text{sin}^2\dfrac{B+C}{2}\\[1em] = \text{sin}^2\dfrac{A}{2} + \text{sin}^2\Big(90° - \dfrac{A}{2}\Big)\\[1em] = \text{sin}^2\dfrac{A}{2} + \text{cos}^2\dfrac{A}{2}\\[1em] = 1$

R.H.S. = 1

∴ L.H.S. = R.H.S.

Hence, $\text{sin}^2\dfrac{A}{2}$ + $\text{sin}^2\dfrac{B+C}{2}$ = 1.

Question 10

If sec (90° - 3A).cos 48° = 1 and 0 ≤ 3A ≤ 90°; find the value of angle A.

Answer

Given:

sec (90° - 3A) . cos 48° = 1

⇒ cosec 3A . cos 48° = 1

⇒ $\dfrac{1}{\text{sin 3A}}$ . cos 48° = 1

⇒ sin 3A = cos 48°

⇒ sin 3A = cos (90° - 42°)

⇒ sin 3A = sin 42°

So, 3A = 42°

⇒ A = $\dfrac{42°}{3}$

⇒ A = 14°

Hence, A = 14°.

Question 11

In △ABC, angle C is 90° then find the value of sin (A + B).

Answer

In △ ABC,

∠ A + ∠ B + ∠ C = 180°

⇒ ∠ A + ∠ B + 90° = 180°

⇒ ∠ A + ∠ B = 180° - 90°

⇒ ∠ A + ∠ B = 90°

⇒ sin(A + B) = sin 90°

⇒ sin(A + B) = 1

Hence, sin (A + B) = 1.

Question 12

If sec A sin A = 0, find the value of cos A.

Answer

sec A sin A = 0

⇒ $\dfrac{1}{\text{cos A}}$ sin A = 0

⇒ tan A = 0

⇒ tan A = tan 0°

Thus, A = 0°.

Now, cos A = cos 0° = 1

Hence, cos A = 1.

Question 13

Find angle A, if sec 2A = cosec (A + 48°).

Answer

We know that,

sec θ = cosec (90° - θ)

Solving,

⇒ sec 2A = cosec (A + 48°)

⇒ cosec (90° - 2A) = cosec (A + 48°)

⇒ 90° - 2A = A + 48°

⇒ A + 2A = 90° - 48°

⇒ 3A = 42°

⇒ A = $\dfrac{42°}{3}$ = 14°.

Hence, A = 14°.

Chapter 21

Trigonometrical Ratios

Class - 9 Concise Mathematics Selina

Exercise 21(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Exercise 21(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Exercise 21(C)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2(i)

Question 2(ii)

Question 2(iii)

Question 2(iv)

Question 2(v)

Question 2(vi)

Question 3(i)

Question 3(ii)

Question 3(iii)

Question 4(i)

Question 4(ii)

Question 4(iii)

Question 4(iv)

Question 4(v)

Question 4(vi)

Question 5(i)

Question 5(ii)

Question 5(iii)

Question 6

Question 7(i)

Question 7(ii)

Question 8(i)

Question 8(ii)

Question 8(iii)

Question 8(iv)

Question 9(i)

Question 9(ii)

Question 9(iii)

Question 10(i)

Question 10(ii)

Question 10(iii)