ML Aggarwal Solutions for Class 9 Maths Chapter 18 Trigonometric Ratios and Standard Angles

ML Aggarwal Solutions for Class 9 Maths Chapter 18 Trigonometric Ratios and Standard Angles provide detailed solutions to the chapter test problems. The solutions are created by individual subject experts at BYJU’S following the latest ICSE guidelines. In order to secure high marks in Mathematics, practising to solve problems on a daily basis is important. Students can access the solutions of the chapter test question from the ML Aggarwal Solutions for Class 9 Maths Chapter 18 Trigonometric Ratios and Standard Angles PDF, which is available in the link attached below.

Chapter 18 contains problems with regard to proving trigonometric identities and solving trigonometric equations using the trigonometric ratios of standard angles. ML Aggarwal solutions are readily available in PDF format so that the students can make use of them offline while solving the exercise problems of ML Aggarwal textbook. With the help of this resource, students will be able to improve their problem-solving skills and also learn to manage time by learning simple tricks and shortcuts embedded in these solutions.

ML Aggarwal Solutions for Class 9 Maths Chapter 18 Trigonometric Ratios and Standard Angles Download PDF

 

ML Aggarwal Solutions for Class 9 Maths Chapter 18 TR and Standard Angles
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Access ML Aggarwal Solutions for Class 9 Maths Chapter 18 Trigonometric Ratios and Standard Angles

Exercise 18.1

1. Find the values of

(i) 7 sin 300 cos 600

(ii) 3 sin2 450 + 2 cos2 600

(iii) cos2 450 + sin2 600 + sin2 300

(iv) cos 900 + cos2 450 sin 300 tan 450.

Solution:

(i) 7 sin 300 cos 600

Substituting the values

= 7 × ½ × ½

= (7 × 1 × 1)/(2 × 2)

= 7/4

(ii) 3 sin2 450 + 2 cos2 600

Substituting the values

= 3 × (1/2)2 + 2 × (1/2)2

By further calculation

= 3 × ½ + 2 × ¼

= 3/2 + ½

So we get

= (3 + 1)/2

= 4/2

= 2

(iii) cos2 450 + sin2 600 + sin2 300

Substituting the values

= (1/2)2 + (√3/2)2 + (1/2)2

By further calculation

= ½ + ¾ + ¼

Taking LCM

= (2 + 3 + 1)/4

= 6/4

= 3/2

(iv) cos 900 + cos2 450 sin 300 tan 450

Substituting the values

= 0 + (1/2)2 × ½ × 1

By further calculation

= ½ × ½ × 1

= ¼

2. Find the values of

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 1

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 2

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 3

(iii) 4/3 tan2 300 + sin2 600 – 3 cos2 600 + ¾ tan2 600 – 2 tan2 450

Substituting the values

= 4/3 (1/√3)2 + (√3/2)2 – 3 (1/2)2 + ¾ × (√3)2 – 2 × 12

By further calculation

= 4/3 × 1/3 + ¾ – 3 × ¼ + ¾ × 3 – 2 × 1

= 4/9 + ¾ – 3/4 + 9/4 – 2

So we get

= 4/9 + 9/4 – 2

Taking LCM

= (16 + 81 – 72)/36

= (97 – 72)/36

= 25/36

3. Find the values of

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 4

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 5

(ii) 2√2 cos 450 cos 600 + 2√3 sin 300 tan 600 – cos 00

Substituting the values

= 2√2 × 1/√2 × ½ + 2√3 × ½ × √3 – 1

By further calculation

= 2 × 1/1 × 1/2 + 2 × 3 × ½ – 1

= 1 + 3 – 1

= 3

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 6

4. Prove that

(i) cos2 300 + sin 300 + tan2 450 = 2 ¼

(ii) 4 (sin4 300 + cos4 600) – 3 (cos2 450 – sin2 900) = 2

(iii) cos 600 = cos2 300 – sin2 300.

Solution:

(i) cos2 300 + sin 300 + tan2 450 = 2 ¼

Consider

LHS = cos2 300 + sin 300 + tan2 450

Substituting the values

= (√3/2)2 + ½ + 12

By further calculation

= ¾ + ½ + 1

Taking LCM

= (3 + 2 + 4)/4

= 9/4

= 2 ¼

= RHS

Therefore, LHS = RHS.

(ii) 4 (sin4 300 + cos4 600) – 3 (cos2 450 – sin2 900) = 2

Consider

LHS = 4 (sin4 300 + cos4 600) – 3 (cos2 450 – sin2 900)

Substituting the values

= 4[(½)4 + (½)4] – 3 [(1/√2)2 – 12]

It can be written as

= 4[½ × ½ × ½ × ½ + ½ × ½ × ½ × ½] – 3 [½ – 1]

By further calculation

= 4 [1/16 + 1/16] – 3 (- ½)

= 4[(1 + 1)/16] + 3/2

So we get

= (4 × 3)/16 + 3/2

= 8/16 + 3/2

= ½ + 3/2

= (1 + 3)/2

= 4/2

= 2

= RHS

Therefore, LHS = RHS.

(iii) cos 600 = cos2 300 – sin2 300

Consider

LHS = cos 600 = ½

RHS = cos2 300 – sin2 300

Substituting the values

= (√3/2)2 + (1/2)2

By further calculation

= ¾ – ¼

= (3 – 1)/4

= 2/4

= ½

= RHS

Therefore, LHS = RHS.

5. (i) If x = 300, verify that tan 2x = 2tanx/ (1- tan2 x).

(ii) If x = 150, verify that 4 sin 2x cos 4x sin 6x = 1.

Solution:

(i) It is given that

x = 300

Consider LHS = tan 2x

Substituting the value of x

= tan 600 

= √3

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 7

Therefore, LHS = RHS.

(ii) It is given that

x = 150

2x = 15 × 2 = 300

4x = 15 × 4 = 600

6x = 15 × 6 = 900

Here

LHS = 4 sin 2x cos 4x sin 6x

It can be written as

= 4 sin 300 cos 600 sin 900

So we get

= 4 × ½ × ½ × 1

= 1

= RHS

Therefore, LHS = RHS.

6. Find the values of

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 8

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 9

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 10

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 11

7. If θ = 300, verify that

(i) sin 2θ = 2 sin θ cos θ

(ii) cos 2θ = 2 cos2 θ – 1

(iii) sin 3θ = 3 sin θ – 4 sin3 θ

(iv) cos 3θ = 4 cos3 θ – 3 cos θ.

Solution:

It is given that θ = 300

(i) sin 2θ = 2 sin θ cos θ

Consider

LHS = sin 2θ

Substituting the value of θ

= sin 2 × 300

= sin 600

= √3/2

RHS = 2 sin θ cos θ

Substituting the value of θ

= 2 sin 300 cos 300

So we get

= 2 × ½ × √3/2

= 1 × √3/2

= √3/2

Therefore, LHS = RHS.

(ii) cos 2θ = 2 cos2 θ – 1

Consider

LHS = cos 2θ

Substituting the value of θ

= cos 2 × 300

= cos 600

= ½

RHS = 2 cos2 θ – 1

Substituting the value of θ

= 2 cos2 300 – 1

So we get

= 2 (√3/2)2 – 1

= 2 × ¾ – 1

= 3/2 – 1

= (3 – 2)/2

= ½

Therefore, LHS = RHS.

(iii) sin 3θ = 3 sin θ – 4 sin3 θ

Consider

LHS = sin 3θ

Substituting the value of θ

= sin 3 × 300

= sin 900

= 1

RHS = 3 sin θ – 4 sin3 θ

Substituting the value of θ

= 3 sin 300 – 4 sin3 300

So we get

= 3 × ½ – 4 × (1/2)3

= 3/2 – 4 × 1/8

= 3/2 – ½

Taking LCM

= (3 – 1)/2

= 2/2

= 1

Therefore, LHS = RHS.

(iv) cos 3θ = 4 cos3 θ – 3 cos θ

Consider

LHS = cos 3θ

Substituting the value of θ

= cos 3 × 300

= cos 900

= 0

RHS = 4 cos3 θ – 3 cos θ

Substituting the value of θ

= 4 cos3 300 – 3 cos 300

So we get

= 4 × (√3/2)3 – 3 × (√3/2)

By further calculation

= 4 × 3√3/8 – 3√3/2

= 3√3/2 – 3√3/2

= 0

Therefore, LHS = RHS.

8. If θ = 300, find the ratio 2 sin θ: sin 2 θ.

Solution:

It is given that θ = 300

We know that

2 sin θ: sin 2θ = 2 sin 300: sin 2 × 300

So we get

= 2 sin 300: sin 600

= 2 sin 300/ sin 600

Substituting the values

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 12

= 2: √3

Therefore, 2 sin θ: sin 2θ = 2: √3.

9. By means of an example, show that sin (A + B) ≠ sin A + sin B.

Solution:

Consider A = 300 and B = 600

LHS = sin (A + B)

Substituting the values of A and B

= sin (300 + 600)

= sin 900

= 1

RHS = sin A + sin B

Substituting the values

= sin 300 + sin 600

So we get

= ½ + √3/2

= (1 + √3)/2

Therefore, LHS ≠ RHS i.e. sin (A + B) ≠ sin A + sin B.

10. If A = 600 and B = 300, verify that

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

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

(iii) sin (A – B) = sin A cos B – cos A sin B

(iv) tan (A – B) = (tan A – tan B)/ (1 + tan A tan B).

Solution:

It is given that A = 600 and B = 300

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

Here

LHS = sin (A + B)

Substituting the values of A and B

= sin (600 + 300)

= sin 900

= 1

RHS = sin A cos B + cos A sin B

Substituting the values of A and B

= sin 600 cos 300 + cos 600 sin 300

So we get

= √3/2 × √3/2 + ½ × ½

By further calculation

= ¾ + ¼

= 4/4

= 1

Therefore, LHS = RHS.

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

Here

LHS = cos (A + B)

Substituting the value of A and B

= cos (600 + 300)

= cos 900

= 0

RHS = cos A cos B – sin A sin B

Substituting the value of A and B

= cos 600 cos 300 – sin 600 sin 300

So we get

= ½ × √3/2 – √3/2 + ½

= √3/4 – √3/4

= 0

Therefore, LHS = RHS.

(iii) sin (A – B) = sin A cos B – cos A sin B

Here

LHS = sin (A – B)

Substituting the values of A and B

= sin (600 – 300)

= sin 300

= ½

RHS = sin A cos B – cos A sin B

Substituting the values of A and B

= sin 600 cos 300 – cos 600 sin 300

So we get

= √3/2 × √3/2 – ½ × ½

= ¾ – ¼

= (3 – 1)/ 4

= 2/4

= ½

Therefore, LHS = RHS.

(iv) tan (A – B) = (tan A – tan B)/ (1 + tan A tan B)

Here

LHS = tan (A – B)

Substituting the values of A and B

= tan (600 – 300)

= tan 300

= 1/√3

RHS = (tan A – tan B)/ (1 + tan A tan B)

Substituting the values of A and B

= (tan 600 – tan 300)/ (1 + tan 600 tan 300)

So we get

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 13

We get

= 2/√3 × ½

= 1/√3

Therefore, LHS = RHS.

11. (i) If 2θ is an acute angle and 2 sin 2θ = √3, find the value of θ.

(ii) If 200 + x is an acute angle and cos (200 + x) = sin 600, then find the value of x.

(iii) If 3 sin2 θ = 2 ¼ and θ is less than 900, find the value of θ.

Solution:

(i) It is given that

2θ is an acute angle

2 sin 2θ = √3

It can be written as

sin 2θ = √3/2 = sin 600

By comparing

2θ = 600

So we get

θ = 600/2 = 300

Therefore, θ = 300.

(ii) It is given that

200 + x is an acute angle

cos (200 + x) = sin 600

It can be written as

cos (200 + x) = sin 600 = cos (900 – 600)

= cos 300

By comparing

200 + x = 300

x = 300 – 200 = 100

Therefore, x = 100.

(iii) It is given that

3 sin2 θ = 2 ¼

θ is less than 900

We can write it as

sin2 θ = 9/(4 × 3) = ¾

So we get

sin θ = √3/2 = sin 600

By comparing

θ = 600

Therefore, θ = 600.

12. If θ is an acute angle and sin θ = cos θ, find the value of θ and hence, find the value of 2 tan2 θ + sin2 θ – 1.

Solution:

It is given that

sin θ = cos θ

We can write it as

sin θ/ cos θ = 1

tan θ = 1

We know that tan 450 = 1

tan θ = tan 450

So we get

θ = 450

We know that

2 tan2 θ + sin2 θ – 1 = 2 tan2 450 + sin2 450 – 1

Substituting the values

= 2 (1)2 + (1/√2)2 – 1

By further calculation

= 2 × 1 × 1 + ½ – 1

= 2 + ½ – 1

= 5/2 – 1

Taking LCM

= (5 – 2)/2

= 3/2

Therefore, 2 tan2 θ + sin2 θ – 1 = 3/2.

13. From the adjoining figure, find

(i) tan x0

(ii) x

(iii) cos x0

(iv) use sin x0 to find y.

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 14

Solution:

(i) tan x0 = perpendicular/base

It can be written as

= AB/BC

= √3/1

= √3

(ii) tan x0 = √3

We know that tan 600 = √3

tan x0 = tan 600

x = 60

(iii) We know that

cos x0 = cos 600

So we get

cos x0 = ½

(iv) sin x0 = perpendicular/hypotenuse = AB/AC

Substitute x = 60 from (ii)

sin 600 = √3/y

We know that sin 600 = √3/2

√3/2 = √3/y

By further calculation

y = (√3 × 2)/ √3

y = (2 × 1)/1 = 2

Therefore, y = 2.

14. If 3θ is an acute angle, solve the following equations for θ:

(i) 2 sin 3θ = √3

(ii) tan 3θ = 1.

Solution:

(i) 2 sin 3θ = √3

It can be written as

sin 3θ = √3/2

We know that sin 600 = √3/2

sin 3θ = sin 600

3θ = 600

So we get

θ = 60/3 = 200

(ii) tan 3θ = 1

We know that tan 450 = 1

tan 3θ = tan 450

So we get

3θ = 450

θ = 150

15. If tan 3x = sin 450 cos 450 + sin 300, find the value of x.

Solution:

We know that

tan 3x = sin 450 cos 450 + sin 300

Substituting the values

= 1/√2 × 1/√2 + ½

= ½ + ½

= 1

We know that

tan 3x = tan 450

By comparing

3x = 450

x = 45/3 = 150

Therefore, the value of x is 150.

16. If 4 cos2 x0 – 1 = 0 and 0 ≤ x ≤ 90, find

(i) x

(ii) sin2 x0 + cos2 x0

(iii) cos2 x0 – sin2 x0.

Solution:

It is given that

4 cos2 x0 – 1 = 0

4cos2 x0 = 1

It can be written as

cos2 x0 = ¼

cos x0 = ± √1/4

cos x0 = + √1/4 [0 ≤ x ≤ 900, then cos x0 is positive]

cos x0 = ½

We know that cos 600 = ½

cos x0 = cos 600

By comparing

x = 60

(ii) sin2 x0 + cos2 x0 = sin2 600 + cos2 600

Substituting the values

= (√3/2)2 + (1/2)2

By further calculation

= ¾ + ¼

= (3 + 1)/4

= 4/4

= 1

Therefore, sin2 x0 + cos2 x0 = 1.

(iii) cos2 x0 – sin2 x0 = cos2 600 – sin2 600

Substituting the values

= (1/2)2 – (√3/2)2

By further calculation

= ¼ – √3/2 × √3/2

= ¼ – ¾

= (1 – 3)/4

= -2/4

= – ½

Therefore, cos2 x0 – sin2 x0 = – ½.

17. (i) If sec θ = cosec θ and 00 ≤ θ ≤ 900, find the value of θ.

(ii) If tan θ = cot θ and 00 ≤ θ ≤ 900, find the value of θ

Solution:

(i) It is given that

sec θ = cosec θ

We know that

sec θ = 1/ cos θ

cosec θ = 1/ sin θ

So we get

1/ cos θ = 1/sin θ

sin θ/ cos θ = 1

tan θ = 1

Here tan 450 = 1

tan θ = tan 450

θ = 450

(ii) It is given that

tan θ = cot θ

We know that cot θ = 1/tan θ

tan θ = 1/ tan θ

So we get

tan2 θ = 1

tan θ = ± √1

tan θ = + 1 [0 ≤ θ ≤ 900, tan θ is positive]

tan θ = tan 450

By comparing

θ = 450

18. If sin 3x = 1 and 00 ≤ 3x ≤ 900, find the values of

(i) sin x

(ii) cos 2x

(iii) tan2 x – sec2 x.

Solution:

It is given that

sin 3x = 1

We know that sin 900 = 1

sin 3x = sin 900

By comparing

3x = 90

x = 90/3

x = 300

(i) sin x = sin 300 = 1/2

(ii) cos 2x = cos 2 × 30 = cos 600 = 1/2

(iii) tan2 x – sec2 x = tan2 300 – sec2 300

Substituting the values

= (1/√3)2 – (2/√3)2

By further calculation

= 1/3 – 4/3

= (1 – 4)/3

= – 3/3

= -1

Therefore, tan2 x – sec2 x = – 1.

19. If 3 tan2 θ – 1 = 0, find cos 2θ, given that θ is acute.

Solution:

It is given that

3 tan2 θ – 1 = 0

We can write it as

3 tan2 θ = 1

tan2 θ = 1/3

tan θ = 1/√3 [θ is acute so tan θ is positive]

θ = 300

So we get

cos 2θ = cos 2 × 300 = cos 600 = ½

20. If sin x + cos y = 1, x = 300 and y is acute angle, find the value of y.

Solution:

It is given that

sin x + cos y = 1

x = 300

Substituting the values

sin 300 + cos y = 1

1/2 + cos y = 1

It can be written as

cos y = 1 – ½

Taking LCM

cos y = (2 – 1)/2 = ½

We know that cos 600 = ½

cos y = cos 600

So we get

y = 600

21. If sin (A + B) = √3/2 = cos (A – B), 00 < A + B ≤ 900 (A > B), find the values of A and B.

Solution:

It is given that

sin (A + B) = √3/2 = cos (A – B)

Consider

sin (A + B) = √3/2

We know that sin 600 = √3/2

sin (A + B) = sin 600

A + B = 600 …… (1)

Similarly

cos (A – B) = √3/2

We know that cos 300 = √3/2

cos (A – B) = cos 300

A – B = 300 ……. (2)

By adding both the equations

A + B + A – B = 600 + 300

So we get

2A = 900

A = 900/2 = 450

Now substitute the value of A in equation (1)

450 + B = 600

By further calculation

B = 600 – 450 = 150

Therefore, A = 450 and B = 150.

22. If the length of each side of a rhombus is 8 cm and its one angle is 600, then find the lengths of the diagonals of the rhombus.

Solution:

It is given that

Each side of a rhombus = 8 cm

One angle = 600

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 15

We know that the diagonals bisect the opposite angles

∠OAB = 600/2 = 300

In right ∠AOB

sin 300 = OB/AB

So we get

½ = OB/8

By further calculation

OB = 8/2 = 4 cm

BD = 2OB = 2 × 4 = 8 cm

cos 300 = AO/AB

Substituting the values

√3/2 = AO/8

By further calculation

AO = 8√3/2 = 4√3

Here

AC = 4√3 × 2 = 8 √3 cm

Therefore, the length of the diagonals of the rhombus are 8 cm and 8√3 cm.

23. In the right-angled triangle ABC, ∠C = 900 and ∠B = 600. If AC = 6 cm, find the lengths of the sides BC and AB.

Solution:

In the right-angled triangle ABC, C = 900 and B = 600

AC = 6 cm

We know that

tan B = AC/BC

Substituting the values

tan 600 = 6/BC

So we get

√3 = 6/BC

BC = 6/√3

It can be written as

= 6√3/(√3 + √3)

= 6√3/3

= 2√3 cm

sin 600 = AC/AB

Substituting the values

√3/2 = 6/AB

By further calculation

AB = (6 × 2)/ √3

So we get

AB = (12 × √3)/(√3 × √3)

= 12√3/3

= 4√3 cm

Therefore, the lengths of the sides BC = 2√3 cm and AB = 4√3 cm.

24. In the adjoining figure, AP is a man of height 1.8 m and BQ is a building 13.8 m high. If the man sees the top of the building by focusing his binoculars at an angle of 300 to the horizontal, find the distance of the man from the building.

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 16

Solution:

It is given that

Height of man AP = 1.8 m

Height of building BQ = 13.8 m

Angle of elevation from the top of the building to the man = 300

Consider AB as the distance of the man from the building

AB = x then PC = x

QC = 13.8 – 1.8 = 12 m

In right △ PQC

tan θ = QC/PC

Substituting the values

tan 300 = 12/x

By further calculation

1/√3 = 12/x

x = 12 √3 m

Therefore, the distance of the man from the building is 12 √3 m.

25. In the adjoining figure, ABC is a triangle in which ∠B = 450 and ∠C = 600. If AD ⟂ BC and BC = 8 m, find the length of the altitude AD.

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 17

Solution:

In triangle ABC

B = 450 and C = 600

AD ⟂ BC and BC = 8 m

In right △ ABD

tan 450 = AD/BD

So we get

1 = AD/BD

AD = BD

In right △ ACD

tan 600 = AD/DC

So we get

√3 = AD/DC

DC = AD/√3

Here

BD + DC = AD + AD/√3

Taking LCM

BC = (√3AD + AD)/ √3

8 = [AD (√3 + 1)]/ √3

By further calculation

AD = 8√3/(√3 + 1)

It can be written as

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 18

= 4 (3 – √3) m

Therefore, the length of the altitude AD is 4 (3 – √3) m.

Exercise 18.2

Without using trigonometric tables, evaluate the following (1 to 5):

1. (i) cos 180/sin 720

(ii) tan 410/cot 490

(iii) cosec 17030’/sec 720 30’

Solution:

(i) cos 180/sin 720

It can be written as

= cos 180/ sin (900 – 180)

So we get

= cos 180/cos 180

= 1

(ii) tan 410/cot490

It can be written as

= tan 410/ cot (900 – 410)

So we get

= tan 410/ tan 410

= 1

(iii) cosec 17030’/sec 720 30’

It can be written as

= cosec 170 30’/ sec (900 – 170 30’)

So we get

= cosec 170 30’/ cosec 170 30’

= 1

2. (i) cot 400/tan 500 – ½ (cos 350/sin 550)

(ii) (sin 490/cos 410)2 + (cos 410/sin 490)2

(iii) sin 720/cos 180 – sec 320/cosec 580

(iv) cos 750/sin 150 + sin 120/ cos 780 – cos 180/sin 720

(v) sin 250/sec 650 + cos 250/ cosec 650.

Solution:

(i) cot 400/tan 500 – ½ (cos 350/sin 550)

It can be written as

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 19

(ii) (sin 490/cos 410)2 + (cos 410/sin 490)2

It can be written as

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 20

So we get

= 12 + 12

= 1 + 1

= 2

(iii) sin 720/cos 180 – sec 320/cosec 580

It can be written as

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 21

= 1 – 1

= 0

(iv) cos 750/sin 150 + sin 120/ cos 780 – cos 180/sin 720

It can be written as

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 22

So we get

= 1 + 1 – 1

= 1

(v) sin 250/sec 650 + cos 250/ cosec 650

It can be written as

= (sin 250 × cos 650) + (cos 250 × sin 650)

By further calculation

= [sin 250 × cos (900 – 250)] + [cos 250 × sin (900 – 250)]

So we get

= (sin 250 × sin 250) + (cos 250 × cos 250)

= sin2 250 + cos2 250

= 1

3. (i) sin 620 – cos 280

(ii) cosec 350 – sec 550.

Solution:

(i) sin 620 – cos 280

It can be written as

= sin (900 – 280) – cos 280

So we get

= cos 280 – cos 280

= 0

(ii) cosec 350 – sec 550

It can be written as

= cosec 350 – sec (900 – 350)

So we get

= cosec 350 – cosec 350

= 0

4. (i) cos2 260 + cos 640 sin 260 + tan 360/ cot 540

(ii) sec 170/cosec 730 + tan 680/cot 220 + cos2 440 + cos2 460.

Solution:

(i) cos2 260 + cos 640 sin 260 + tan 360/ cot 540

It can be written as

= cos2 260 + cos (900 – 260) sin 260 + tan 360/cot (900 – 360)

We know that

cos (900 – θ) = sin θ

cot (900 – θ) = tan θ

sin2 θ + cos2 θ = 1

So we get

= cos2 260 + sin2 260 + tan 360/tan 360

= 1 + 1

= 2

(ii) sec 170/cosec 730 + tan 680/cot 220 + cos2 440 + cos2 460

It can be written as

= sec (900 – 730)/ cosec 730 + tan (900 – 220)/ cot 220 + cos2 (900 – 460) + cos2 460

By further calculation

= cosec 730/ cosec 730 + cot 220/ cot 220 + sin2 460 + cos2 460

We know that sin2 θ + cos2 θ = 1

So we get

= 1 + 1 + 1

= 3

5. (i) cos 650/sin 250 + cos 320/sin 580 – sin 280 sec 620 + cosec2 300

(ii) sec 290/ cosec 610 + 2 cot 80 cot 170 cot 450 cot 730 cot 820 – 3 (sin2 380 + sin2 520).

Solution:

(i) cos 650/sin 250 + cos 320/sin 580 – sin 280 sec 620 + cosec2 300

It can be written as

= cos 650/ sin (900 – 650) + cos 320/ sin (900 – 320) – sin 280 sec (900 – 280) + cosec2 300

By further calculation

= cos 650/cos 650 + cos 320/ cos 320 – sin 280 cosec 280 + cosec2 300

We know that cosec 300 = 2

= 1 + 1 – 1 + 4

= 5

(ii) sec 290/ cosec 610 + 2 cot 80 cot 170 cot 450 cot 730 cot 820 – 3 (sin2 380 + sin2 520)

It can be written as

= sec 290/ cosec (900 – 290) + 2 cot 80 cot 170 cot 450 cot (900 – 170) cot (900 – 80) – 3 [sin2 380 + sin2 (900 – 380)]

By further calculation

= sec 290/ sec 290 + 2 cot 80 cot 170 × 1 × tan 170 tan 80 – 3 (sin2 380 + cos2 380)

So we get

= 1 + 2 cot 80 tan 80 cot 170 tan 170 × 1 – 3 × 1

We know that

cosec (900 – θ) = sec θ

cot (900 – θ) = tan θ

sin2 θ + cos2 θ = 1

Here

= 1 + 2 × 1 × 1 × 1 – 3

= 1 + 2 – 3

= 0

6. Express each of the following in terms of trigonometric ratios of angles between 00 to 450:

(i) tan 810 + cos 720

(ii) cot 490 + cosec 870.

Solution:

(i) tan 810 + cos 720

It can be written as

= tan (900 – 90) + cos (900 – 180)

So we get

= cot 90 + sin 180

(ii) cot 490 + cosec 870

It can be written as

= cot (900 – 410) + cosec (900 – 30)

So we get

= tan 410 + sec 30

Without using trigonometric tables, prove that (7 to 11):

7. (i) sin2 280 – cos2 620 = 0

(ii) cos2 250 + cos2 650 = 1

(iii) cosec2 670 – tan2 230 = 1

(iv) sec2 220 – cot2 680 = 1.

Solution:

(i) sin2 280 – cos2 620 = 0

Consider

LHS = sin2 280 – cos2 620

It can be written as

= sin2 280 – cos2 (900 – 280)

So we get

= sin2 280 – sin2 280

= 0

= RHS

(ii) cos2 250 + cos2 650 = 1

Consider

LHS = cos2 250 + cos2 650

It can be written as

= cos2 250 + cos2 (900 – 250)

We know that sin2 θ + cos2 θ = 1

So we get

= cos2 250 + sin2 250

= 1

(iii) cosec2 670 – tan2 230 = 1

Consider

LHS = cosec2 670 – tan2 230

It can be written as

= cosec2 670 – tan2 (900 – 670)

We know that cosec2 θ – cot2 θ = 1

So we get

= cosec2 670 – cot2 670

= 1

(iv) sec2 220 – cot2 680 = 1

Consider

LHS = sec2 220 – cot2 680

It can be written as

= sec2 220 – cot2 (900 – 220)

We know that sec2 θ – tan2 θ = 1

So we get

= sec2 220 – tan2 220

= 1

8. (i) sin 630 cos 270 + cos 630 sin 270 = 1

(ii) sec 310 sin 590 + cos 310 cosec 590 = 2.

Solution:

(i) sin 630 cos 270 + cos 630 sin 270 = 1

Consider

LHS = sin 630 cos 270 + cos 630 sin 270

It can be written as

= sin 630 cos (900 – 630) + cos 630 sin (900 – 630)

= sin 630 sin 630 + cos 630 cos 630

We know that sin2 θ + cos2 θ = 1

So we get

= sin2 630 + cos2 630

= 1

(ii) sec 310 sin 590 + cos 310 cosec 590 = 2

Consider

LHS = sec 310 sin 590 + cos 310 cosec 590

It can be written as

= 1/ cos 310 × sin 590 + cos 310 × 1/ sin 590

By further calculation

= sin 590/cos (900 – 590) + cos 310/ sin (900 – 310)

So we get

= sin 590/sin 590 + cos 310/cos 310

= 1 + 1

= 2

= RHS

9. (i) sec 700 sin 200 – cos 200 cosec 700 = 0

(ii) sin2 200 + sin2 700 – tan2 450 = 0.

Solution:

(i) sec 700 sin 200 – cos 200 cosec 700 = 0

Consider

LHS = sec 700 sin 200 – cos 200 cosec 700

By further simplification

= sin 200/ cos 700 – cos 200/ sin 700

It can be written as

= sin 200/ cos (900 – 200) – cos 200/ sin (900 – 200)

So we get

= sin 200/ sin 200 – cos 200/ cos 200

= 1 – 1

= 0

= RHS

(ii) sin2 200 + sin2 700 – tan2 450 = 0

Consider

LHS = sin2 200 + sin2 700 – tan2 450

It can be written as

= sin2 200 + sin2 (900 – 200) – tan2 450

By further calculation

= sin2 200 + cos2 200 – tan2 450

We know that sin2 θ + cos2 θ = 1 and tan 450 = 1

So we get

= 1 – 1

= 0

= RHS

10. (i) cot 540/tan 360 + tan 200/cot 700 – 2 = 0

(ii) sin 500/cos 400 + cosec 400/sec 500 – 4 cos 500 cosec 400 + 2 = 0.

Solution:

(i) cot 540/tan 360 + tan 200/cot 700 – 2 = 0

Consider

LHS = cot 540/tan 360 + tan 200/cot 700 – 2

It can be written as

= cot 540/tan (900 – 540) + tan 200/cot (900 – 200) – 2

By further calculation

= cot 540/ cot 540 + tan 200/ tan 200 – 2

So we get

= 1 + 1 – 2

= 0

= RHS

(ii) sin 500/cos 400 + cosec 400/sec 500 – 4 cos 500 cosec 400 + 2 = 0

Consider

LHS = sin 500/cos 400 + cosec 400/sec 500 – 4 cos 500 cosec 400 + 2

It can be written as

= sin 500/cos (900 – 500) + cosec 400/sec (900 – 400) – 4 cos 500 cosec (900 – 500) + 2

By further calculation

= sin 500/sin 500 + cosec 400/cosec 400 – cos 500 sec 500 + 2

So we get

= 1 + 1 – 4 cos500/cos 500 + 2

= 1 + 1 – 4 + 2

= 4 – 4

= 0

= RHS

11. (i) cos 700/sin 200 + cos 590/sin 310 – 8 sin2 300 = 0

(ii) cos 800/sin 100 + cos 590 cosec 310 = 2.

Solution:

(i) cos 700/sin 200 + cos 590/sin 310 – 8 sin2 300 = 0

Consider

LHS = cos 700/sin 200 + cos 590/sin 310 – 8 sin2 300

It can be written as

= cos 700/sin (900 – 700) + cos 590/sin (900 – 590) – 8 sin2 300

We know that sin 300 = ½

= cos 700/cos 700 + cos 590/ cos 590 – 8 (1/2)2

By further calculation

= 1 + 1 – 8 × ¼

So we get

= 2 – 2

= 0

= RHS

(ii) cos 800/sin 100 + cos 590 cosec 310 = 2

Consider

LHS = cos 800/sin 100 + cos 590 cosec 310

It can be written as

= cos 800/sin (900 – 800) + cos 590/ sin 310

By further simplification

= cos 800/ cos 800 + cos 590/ sin (900 – 590)

So we get

= 1 + cos 590/ cos 590

= 1 + 1

= 2

= RHS

12. Without using trigonometrical tables, evaluate:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 23

(iii) sin2 340 + sin2 560 + 2 tan 180 tan 720 – cot2 300.

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 24

So we get

= 2 + 1 – 3

= 0

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 25

We know that sin2 θ + cos2 θ = 1 and cosec2 θ – cot2 θ = 1

= 1/1

= 1

(iii) sin2 340 + sin2 560 + 2 tan 180 tan 720 – cot2 300

It can be written as

= sin2 340 + [sin (900 – 340)]2 + 2 tan 180 tan (900 – 180) – cot2 300

By further simplification

= sin2 340 + cos2 340 + 2 tan 180 cot 180 – (√3)2

So we get

= 1 + 2 tan 180 × 1/tan 180 – 3

= 1 + 2 – 3

= 0

13. Prove the following:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 26

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 27

We know that

LHS =
ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 28

So we get

= cos θ/ cos θ + sin θ/ sin θ

= 1 + 1

= 2

= RHS

(ii) cos θ sin (900 – θ) + sin θ cos (900 – θ) = 1

Consider

LHS = cos θ sin (900 – θ) + sin θ cos (900 – θ)

It can be written as

= cos θ. cos θ + sin θ. sin θ

So we get

= cos2 θ + sin2 θ

= 1

= RHS

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 29

Consider

LHS =
ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 30

By further calculation

= tan θ/ cot θ + cos θ/ cos θ

So we get

= tan θ × tan θ + 1

= tan2 θ + 1

= sec2 θ

= RHS

14. Prove the following:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 31

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 32

Consider

LHS =
ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 33

It can be written as

= sin A cos A/ cot A

So we get

= (sin A cos A × sin A)/ cos A

= sin2 A

= 1 – cos2 A

= RHS

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 34

Consider

LHS =
ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 35

It can be written as

= cos A/ sec A + sin A/ cosec A

So we get

= cos A × cos A + sin A × sin A

= cos2 A + sin2 A

= 1

= RHS

15. Simplify the following:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 36

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 37

It can be written as

= cos θ/ cos θ + sin θ/ cosec θ – 3 tan2 300

By further calculation

= 1 + sin θ × sin θ – 3 (1/ √3)2

So we get

= sin2 θ + 1 – 3 × 1/3

= sin2 θ + 1 – 1

= sin2 θ +

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 38

It can be written as

= (sec θ cos θ tan θ)/ (sin θ cosec θ tan θ) + cot θ/ cot θ

So we get

= sec θ cos θ/ sin θ cosec θ + 1

= 1/1 + 1

= 1 + 1

= 2

16. Show that

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 39

Solution:

Consider

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 40

We know that cos (900 – θ) = sin θ, tan (900 – θ) = cot θ and tan θ cot θ = 1

So we get

= 1/1

= 1

= RHS

17. Find the value of A if

(i) sin 3A = cos (A – 60), where 3A and A – 60 are acute angles

(ii) tan 2A = cot (A – 180), where 2A and A – 180 are acute angles

(iii) If sec 2A = cosec (A – 270) where 2A is an acute angle, find the measure of ∠A.

Solution:

(i) sin 3A = cos (A – 60), where 3A and A – 60 are acute angles

It is given that

sin 3A = cos (A – 60)

We know that cos (900 – θ) = sin θ

cos (900 – 3A) = cos (A – 60)

By comparing both

900 – 3A = A – 60

By further calculation

900 + 60 = A + 3A

960 = 4A

So we get

A = 960/4 = 240

Hence, the value of A is 240.

(ii) tan 2A = cot (A – 180)

We know that cot (900 – θ) = tan θ

cot (900 – 2A) = cot (A – 180)

By comparing both

900 – 2A = A – 180

By further calculation

900 + 180 = A + 2A

So we get

3A = 1080

A = 1080/3 = 360

Hence, the value of A is 360.

(iii) sec 2A = cosec (A – 270)

We know that cosec (900 – θ) = sec θ

cosec (900 – 2A) = cos (A – 270)

By comparing both

900 – 2A = A – 270

By further calculation

900 + 270 = A + 2A

So we get

3A = 1170

A = 1170/3 = 390

Hence, the value of A is 390.

18. Find the value of θ (00 < θ < 900) if:

(i) cos 630 sec (900 – θ) = 1

(ii) tan 350 cot (900 – θ) = 1.

Solution:

(i) cos 630 sec (900 – θ) = 1

It can be written as

cos 630 = 1/sec (900 – θ)

We know that 1/sec θ = cos θ

cos 630 = cos (900 – θ)

By comparing both

900 – θ = 630

By further calculation

θ = 900 – 630 = 270

(ii) tan 350 cot (900 – θ) = 1

It can be written as

tan 350 = 1/cot (900 – θ)

We know that 1/cot θ = cos θ

tan 350 = tan (900 – θ)

By comparing both

350 = 900 – θ

By further calculation

θ = 900 – 350 = 550

19. If A, B and C are the interior angles of a △ ABC, show that

(i) cos (A + B)/2 = sin C/2

(ii) tan (C + A)/2 = cot B/2.

Solution:

A, B and C are the interior angles of a △ ABC

It can be written as

∠A + ∠B + ∠C = 1800

Dividing both sides by 2

(∠A + ∠B + ∠C)/2 = 1800/2

A/2 + B/2 + C/2 = 900

(i) cos (A + B)/2 = sin C/2

We can write it as

(A + B)/2 = 900 – C/2

We know that

cos (900 – C/2) = sin C/2

Here cos (900 – θ) = sin θ

sin C/2 = sin C/2

(ii) tan (C + A)/2 = cot B/2

We know that (A + C)/2 = 900 – B/2

= tan (900 – B/2)

So we get

= cot B/2

= RHS

Chapter Test

1. Find the values of:
(i) sin2 60° – cos2 45° + 3tan2 30°
(ii) 

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 41
(iii) sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°

Solution:

(i) sin2 60° – cos2 45° + 3tan2 30°

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 42

Therefore, sin2 60° – cos2 45° + 3tan2 30° = 1¼

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 43

Hence,
ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 44= 1

(iii) sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 45

= 2 + 1 + ½ = 3 + ½ = (6 + 1)/2

= 7/2 = 3½

Thus, sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60° = 3½

2. Taking A = 30°, verify that
(i) cos4 A – sin4 A = cos 2A
(ii) 4cos A cos (60° – A) cos (60° + A) = cos 3 A.

Solution:



(i) cos4 A – sin4 A = cos 2A

Let’s take A = 30°

so, we have

L.H.S.= cos4 A – sin4 A = cos4 30° – sin4 30°

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 46

Now,

R.H.S. = cos 2A = cos 2(30o) = ½

Therefore, L.H.S. = R.H.S. hence verified.


(ii) 4 cos A cos (60°- A) cos (60° + A) = cos 3 A

Let’s take A = 30°

L.H.S. = 4 cos A cos (60° – A) cos (60° + A)

= 4 cos 30° cos (60° – 30°) cos (60° + 30°)

= 4 cos 30° cos 30° cos 90°

= 4 × (√3/2) × (√3/2) × 0

= 0

Now,

R.H.S. = cos 3A

= cos (3 × 30°) = cos 90° = 0

Hence, L.H.S. = R.H.S. hence verified.

3. If A = 45° and B = 30°, verify that sin A/ (cos A + sin A + sin B) = 2/3

Solution:

Taking,

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 47

Hence verified.

4. Taking A = 60° and B = 30°, verify that

(i) sin (A + B)/ cos A cos B = tan A + tan B

(ii) sin (A – B)/ sin A sin B = cot B – cot A

Solution:

(i) Here, A = 60° and B = 30°

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 48

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 49

5. If √2 tan 2θ = √6 and θ° < 2θ < 90°, find the value of sin θ + √3 cos θ – 2 tan2 θ.

Solution:

Given,

√2 tan 2θ = √6

tan 2θ = √6/ √2

= √3

= tan 60o

⇒ 2θ = 60o

θ = 30o

Now,

sin θ + √3 cos θ – 2 tan2 θ

= sin 30o + √3 cos 30o – 2 tan2 30o

= ½ + √3 x √3/2 – 2 (1/√3)2

= ½ + 3/2 – 2/3

= 4/2 – 2/3

= (12 – 4)/6

= 8/6

= 4/3

6. If 3θ is an acute angle, solve the following equations for θ:
(i) (cosec 3θ – 2) (cot 2θ – 1) = 0

(ii) (tan θ – 1) (cosec 3θ – 1) = 0

Solution:

(i) (cosec 3θ – 2) (cot 2θ – 1) = 0

Now, either

cosec 3θ – 2 or cot 2θ – 1 = 0

⇒ cosec 3θ = 2 or cot 2θ = 1

So,

cosec 3θ = cosec 3θ° or cot 2θ =cot 45°

⇒ 3θ = 30° or 2θ = 45°

Thus, θ = 30° or 45°.

(ii) (tan θ – 1) (cosec 3θ – 1) = 0

Now, either

tan θ – 1 = 0 or cosec 3θ – 1 = 0

⇒ tan θ = 1 or cosec 3θ = 1

So,

tan θ = tan45° or cosec 3 θ = cosec 90°

⇒ θ = 45° or 3θ = 90° i.e. θ = 30°

Thus, θ = 45° or 30°.

7. If tan (A + B) = √3 and tan (A – B) = 1 and A, B (B < A) are acute angles, find the values of A and B.

Solution:

Given, tan (A + B) = √3

So, tan (A + B) = tan 60° [Since, tan 60° = √3]
⇒ A + B = 60° ……….. (i)

Also, given

tan (A – B) = 1

So, tan (A – B) = tan 45° [tan 45° = 1]
⇒ A – B = 45° ………….. (ii)

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

A + B = 60°

A – B = 45°

—————-

2A = 105o

A = 52½o

Now, on substituting the value of A in equation (i), we get

52½o + B = 60°

B = 60° – 52½o = 7½o

Therefore, the value of A = 52½o and B = 7½o

8. Without using trigonometrical tables, evaluate the following:
(i) sin2 28° + sin2 62° – tan2 45°
(ii) 

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 50
(iii) cos 18° sin 72° + sin 18° cos 72°
(iv) 5 sin 50° sec 40° – 3 cos 59° cosec 31°

Solution:

(i) sin2 28° + sin2 62° – tan2 45°

= sin2 28° + sin2 (90° – 28°) – tan2 45°

= sin2 28° + cos2 28° – tan2 45°

= 1 – (1)2  (∵ sin2 θ + cos2 θ = 1 and tan 45° = 1)

= 1 – 1

= 0

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 51

= 2 × 1 + 1 + 1

= 2 + 1 + 1

= 4



(iii) cos 18° sin 12° + sin 18° cos 12°

= cos (90° – 12°) sin 72° + sin (90° – 12°) cos 12°

= sin 72°.sin 12° + cos 12° cos 12°

= sin2 12° + cos2 12°

= 1 (∵ sin2 θ + cos2 θ = 1)



(iv) 5 sin 50° sec 40° – 3 cos 59° cosec 31°

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 52

= 5 – 3

= 2

9. Prove that:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 53

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 54

Thus, L.H.S. = R.H.S.

Hence proved.

10. When 0° < A < 90°, solve the following equations:
(i) sin 3A = cos 2A
(ii) tan 5A = cot A

Solution:

(i) sin 3A = cos 2A

⇒ sin 3A = sin (90° – 2A)

So,

3A = 90° – 2A

3 A + 2A = 90°

5A = 90°

∴ A = 90°/5 = 18°


(ii) tan 5A = cot A

⇒ tan 5A = tan (90° – A)

So,

5A = 90°- A

5A + A = 90°

6A = 90°

∴ A = 90°/6 = 15°

11. Find the value of θ if
(i) sin (θ + 36°) = cos θ, where θ and θ + 36° are acute angles.
(ii) sec 4θ = cosec (θ – 20°), where 4θ and θ – 20° are acute angles.

Solution:

(i) Given, θ and (θ + 36°) are acute angles

And,

sin (θ + 36°) = cos θ = sin (90° – θ) [As, sin (90° – θ) = cos θ]
On comparing, we get

θ + 36° = 90° – θ

θ + θ = 90° – 36°

2θ = 54°

θ = 54°/2

∴ θ = 27°

(ii) Given, θ and (θ – 20°) are acute angles

And,

sec 4θ = cosec (θ – 20°)

cosec (90° – 4θ) = cosec (θ – 20°) [Since, cosec (90° – θ) = sec θ]
On comparing, we get

90° – 4θ = θ – 20°

90° + 20° = θ + 4θ

5θ = 110°

θ = 110°/5

∴ θ = 22°

12. In the adjoining figure, ABC is right-angled triangle at B and ABD is right angled triangle at A. If BD ⊥ AC and BC = 2√3cm, find the length of AD.

ML Aggarwal Solutions for Class 9 Maths Chapter 18 - Image 55

Solution:



Given, ∆ABC and ∆ABD are right angled triangles in which ∠A = 90° and ∠B = 90°

And,

BC = 2√3 cm. AC and BD intersect each other at E at right angle and ∠CAB = 30°.

Now in right ∆ABC, we have

tan θ = BC/AB

⇒ tan 30° = 2√3/ AB

⇒ 1/√3 = 2√3/ AB

⇒ AB = 2√3 × √3 = 2 × 3 = 6 cm.

In ∆ABE, ∠EAB = 30° and ∠EAB = 90°

Hence,

∠ABE or ∠ABD = 180° – 90° – 30°

= 60°

Now in right ∆ABD, we have

tan 60° = AD/AB 

⇒ √3 = AD/6

Thus, AD = 6√3 cm.

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