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

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

Here, we are providing the solutions for Chapter 18 Trigonometric Ratios and Standard Angles from ML Aggarwal Textbook for Class 9 ICSE Mathematics. Solutions of the eighteen chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 9 Chapter 18 Trigonometric Ratios and Standard Angles ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on finding the value of trigonometric ratios for different angles and showing different trigonometric ratios conversion from one to another. 

Exercise 18.1

1. Find the values of

(i) 7 sin 30° cos 60°

(ii) 3 sin² 45° + 2 cos² 60°

(iii) cos² 45° + sin² 60° + sin² 30°

(iv) cos 90° + cos² 45° sin 30° tan 45°.

Solution

(i) 7 sin 30° cos 60°

Substituting the values

= 7 ×½ ×½

= (7×1×1)/(2×2)

= 7/4

(ii) 3 sin² 45° + 2 cos² 60°

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 45° + sin² 60° + sin² 30°

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 90° + cos² 45° sin 30° tan 45°

Substituting the values

= 0 + (1/√2)² × ½ × 1

By further calculation

= ½ × ½ × 1

= ¼

2. Find the values of

(i) (sin²45° + cos²45°)/tan²60°

(ii) (sin30° - sin90° + 2cos0°)/tan² 60°

(iii) 4/3 tan²30° + sin²60° - 3cos²60° + 3/4 tan²60° - 2tan²45°

Solution

(iii) 4/3 tan²30° + sin²60° - 3cos²60° + ¾tan²60° - 2tan²45°

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

(i) (sin30°/cos²45°) - 3tan30° + 5cos90°

(ii) 2√2 cos45° cos60° + 2√3 sin30° tan60° - cos30°

(iii) 4/5 tan²60° - (2/sin²30°) - 3/4 tan²30°

Solution

(ii) 2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°

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

4. Prove that

(i) cos² 30° + sin 30° + tan² 45° = 2 ¼

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

(iii) cos 60° = cos² 30° – sin² 30°.

Solution

(i) cos² 30° + sin 30° + tan² 45° = 2 ¼

Consider

LHS = cos² 30° + sin 30° + tan² 45°

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 30° + cos⁴ 60°) – 3 (cos² 45° – sin² 90°) = 2

Consider

LHS = 4 (sin⁴ 30° + cos⁴ 60°) – 3 (cos² 45° – sin² 90°)

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 60° = cos² 30° – sin² 30°

Consider

LHS = cos 60° = ½

RHS = cos² 30° – sin² 30°

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 = 30°, verify that tan 2x = 2tanx/ (1- tan² x).

(ii) If x = 15°, verify that 4 sin 2x cos 4x sin 6x = 1.

Solution

(i) It is given that

x = 30°

Consider LHS = tan 2x

Substituting the value of x

= tan 60°

= √3

Therefore, LHS = RHS.

(ii) It is given that

x = 15°

2x = 15 × 2 = 30°

4x = 15 × 4 = 60°

6x = 15 × 6 = 90°

Here

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

It can be written as

= 4 sin 30° cos 60° sin 90°

So we get

= 4 × ½ × ½ × 1

= 1

= RHS

Therefore, LHS = RHS.

6. Find the values of

Solution

7. If θ = 30°, 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 = 30°

(i) sin 2θ = 2 sin θ cos θ

Consider

LHS = sin 2θ

Substituting the value of θ

= sin 2 × 30°

= sin 60°

= √3/2

RHS = 2 sin θ cos θ

Substituting the value of θ

= 2 sin 30° cos 30°

So we get

= 2 × ½ × √3/2

= 1 × √3/2

= √3/2

Therefore, LHS = RHS.

(ii) cos 2θ = 2 cos²θ – 1

Consider

LHS = cos 2θ

Substituting the value of θ

= cos 2 × 30°

= cos 60°

= ½

RHS = 2 cos² θ – 1

Substituting the value of θ

= 2 cos² 30° – 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 × 30°

= sin 90°

= 1

RHS = 3 sin θ – 4 sin³ θ

Substituting the value of θ

= 3 sin 30° – 4 sin³ 30°

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 × 30°

= cos 90°

= 0

RHS = 4 cos³ θ – 3 cos θ

Substituting the value of θ

= 4 cos³ 30° – 3 cos 30°

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 = 30°, find the ratio 2 sin θ: sin 2 θ.

Solution

It is given that θ = 30°

We know that

2 sin θ: sin 2θ = 2sin 30°: sin(2×30°)

So we get

= 2 sin 30°: sin 60°

= 2 sin 30°/sin 60°

Substituting the values

= 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 = 30° and B = 60°

LHS = sin (A + B)

Substituting the values of A and B

= sin (30° + 60°)

= sin 90°

= 1

RHS = sin A + sin B

Substituting the values

= sin 30° + sin 60°

So we get

= ½ + √3/2

= (1 + √3)/2

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

10. If A = 60° and B = 30°, 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 = 60° and B = 30°

(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 (60° + 30°)

= sin 90°

= 1

RHS = sin A cos B + cos A sin B

Substituting the values of A and B

= sin 60° cos 30° + cos 60° sin 30°

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 (60° + 30°)

= cos 90°

= 0

RHS = cos A cos B – sin A sin B

Substituting the value of A and B

= cos 60° cos 30° – sin 60° sin 30°

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 (60° – 30°)

= sin 30°

= ½

RHS = sin A cos B – cos A sin B

Substituting the values of A and B

= sin 60° cos 30° – cos 60° sin 30°

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 (60° – 30°)

= tan 30°

= 1/√3

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

Substituting the values of A and B

= (tan 60° – tan 30°)/(1 + tan 60° tan 30°)

So we get

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 20° + x is an acute angle and cos (20° + x) = sin 60°, then find the value of x.

(iii) If 3 sin² θ = 2 ¼ and θ is less than 90°, 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 60°

By comparing

2θ = 60°

So we get

θ = 60°/2 = 30°

Therefore, θ = 30°

(ii) It is given that

20° + x is an acute angle

cos (20° + x) = sin 60°

It can be written as

cos (20° + x) = sin 60° = cos (90° – 60°)

= cos 30°

By comparing

20° + x = 30°

⇒ x = 30° – 20° = 10°

Therefore, x = 10⁰.

(iii) It is given that

3 sin² θ = 2 ¼

θ is less than 90°

We can write it as

sin² θ = 9/(4 × 3) = ¾

So we get

sin θ = √3/2 = sin 60°

By comparing

θ = 60°

Therefore, θ = 60°.

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 45° = 1

tan θ = tan 45°

So we get

θ = 45°

We know that

2 tan² θ + sin² θ – 1 = 2 tan² 45° + sin² 45° – 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 x°

(ii) x

(iii) cos x°

(iv) use sin x° to find y.

Solution

(i) tan x° = perpendicular/base

It can be written as

= AB/BC

= √3/1

= √3

(ii) tan x° = √3

We know that tan 60° = √3

tan x° = tan 60°

x = 60

(iii) We know that

cos x° = cos 60°

So we get

cos x° = ½

(iv) sin x° = perpendicular/hypotenuse = AB/AC

Substitute x = 60 from (ii)

sin 60° = √3/y

We know that sin 60° = √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 60° = √3/2

sin 3θ = sin 60°

⇒ 3θ = 60°

So we get

θ = 60/3 = 20°

(ii) tan 3θ = 1

We know that tan 45° = 1

tan 3θ = tan 45°

So we get

3θ = 45°

⇒ θ = 15°

15. If tan 3x = sin 45° cos 45° + sin 30°, find the value of x.

Solution

We know that

tan 3x = sin 45° cos 45° + sin 30°

Substituting the values

= (1/√2 × 1/√2) + ½

= ½ + ½

= 1

We know that

tan 3x = tan 45°

By comparing

3x = 45°

x = 45/3 = 15°

Therefore, the value of x is 15°.

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

(i) x

(ii) sin² x° + cos² x°

(iii) cos² x° – sin² x°

Solution

It is given that

4 cos² x° – 1 = 0

⇒ 4cos² x° = 1

It can be written as

cos² x° = ¼

⇒ cos x° = ± √1/4

⇒ cos x° = + √1/4 [0 ≤ x ≤ 90°, then cos x° is positive]

⇒ cos x° = ½

We know that cos 60° = ½

cos x° = cos 60°

By comparing

x = 60

(ii) sin² x° + cos² x° = sin² 60° + cos² 60°

Substituting the values

= (√3/2)² + (1/2)²

By further calculation

= ¾ + ¼

= (3 + 1)/4

= 4/4

= 1

Therefore, sin² x° + cos² x° = 1.

(iii) cos² x° – sin² x° = cos² 60° – sin² 60°

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 x° – sin² x° = – ½.

17. (i) If sec θ = cosec θ and 0° ≤ θ ≤ 90°, find the value of θ.

(ii) If tan θ = cot θ and 0° ≤ θ ≤ 90°, 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 45° = 1

tan θ = tan 45°

θ = 45°

(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 ≤ 90°, tan θ is positive]

⇒ tan θ = tan 45°

By comparing

θ = 45°

18. If sin 3x = 1 and 0° ≤ 3x ≤ 90°, 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 90° = 1

sin 3x = sin 90°

By comparing

3x = 90

⇒ x = 90/3

⇒ x = 30°

(i) sin x = sin 30° = 1/2

(ii) cos 2x = cos 2 × 30 = cos 60° = 1/2

(iii) tan² x – sec² x = tan² 30° – sec² 30°

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 = 30°

So we get

cos 2θ = cos 2 × 30° = cos 60° = ½

20. If sin x + cos y = 1, x = 30° and y is acute angle, find the value of y.

Solution

It is given that

sin x + cos y = 1

x = 30°

Substituting the values

sin 30° + 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 60° = ½

cos y = cos 60°

So we get

y = 60°

21. If sin (A + B) = √3/2 = cos (A – B), 0° < A + B ≤ 90° (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 60 = √3/2

sin (A + B) = sin 60°

A + B = 60° ...(1)

Similarly

cos (A – B) = √3/2

We know that cos 30° = √3/2

cos (A – B) = cos 30°

A – B = 30° …(2)

By adding both the equations

A + B + A – B = 60° + 30°

So we get

2A = 90°

⇒ A = 90°/2 = 45°

Now substitute the value of A in equation (1)

45° + B = 60°

By further calculation

B = 60° – 45° = 15°

Therefore, A = 45° and B = 15°.

22. If the length of each side of a rhombus is 8 cm and its one angle is 60°, then find the lengths of the diagonals of the rhombus.

Solution

It is given that

Each side of a rhombus = 8 cm

One angle = 60°

We know that the diagonals bisect the opposite angles

∠OAB = 60°/2 = 30°

In right ∠AOB

sin 30° = OB/AB

So we get

½ = OB/8

By further calculation

OB = 8/2 = 4 cm

BD = 2OB = 2 × 4 = 8 cm

cos 30° = 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 = 90° and ∠B = 60°. If AC = 6 cm, find the lengths of the sides BC and AB.

Solution

In the right-angled triangle ABC, ∠C = 90° and ∠B = 60°

AC = 6 cm

We know that

tan B = AC/BC

Substituting the values

tan 60° = 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 60° = 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 30° to the horizontal, find the distance of the man from the building.

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 = 30°

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 30° = 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 = 45° and ∠C = 60°. If AD ⟂ BC and BC = 8 m, find the length of the altitude AD.

Solution

In triangle ABC

∠B = 45° and ∠C = 60°

AD ⟂ BC and BC = 8 m

In right △ ABD

tan 45° = AD/BD

So we get

1 = AD/BD

AD = BD

In right △ ACD

tan 60° = 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

= 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 18°/sin 72°

(ii) tan 41°/cot 49°

(iii) cosec 17°30’/sec 72°30’

Solution

(i) cos 18°/sin 72°

It can be written as

= cos 18°/sin (90° – 18°)

So we get,

= cos 18°/cos 18°

= 1

(ii) tan 41°/cot49°

It can be written as

= tan 41°/cot (90° – 41°)

So we get

= tan 41°/tan 41°

= 1

(iii) cosec 17°30’/sec 72°30’

It can be written as

= cosec 17°30’/sec (90° – 17°30’)

So we get

= cosec 17°30’/cosec 17°30’

= 1

2. (i) cot 40°/tan 50° – ½ (cos 35°/sin 55°)

(ii) (sin 49°/cos 41°)² + (cos 41°/sin 49°)²

(iii) sin 72°/cos 18° – sec 32°/cosec 58°

(iv) cos 75°/sin 15° + sin 12°/cos 78°– cos 18°/sin 72°

(v) sin 25°/sec 65° + cos 25°/ cosec 65°.

Solution

(i) cot 40°/tan 50° – ½ (cos 35°/sin 55°)

It can be written as

(ii) (sin 49°/cos 41°)² + (cos 41°/sin 49°)²

It can be written as

So we get

= 1² + 1²

= 1 + 1

= 2

(iii) sin 72°/cos 18° – sec 32°/cosec 58°

It can be written as

= 1 – 1

= 0

(iv) cos 75°/sin 15° + sin 12°/ cos 78° – cos 18°/sin 72°

It can be written as

So we get

= 1 + 1 – 1

= 1

(v) sin 25°/sec 65° + cos 25°/ cosec 65°

It can be written as

= (sin 25° × cos 65°) + (cos 25° × sin 65°)

By further calculation

= [sin 25° × cos (90°– 25°)] + [cos 25° × sin (90° – 25°)]

So we get

= (sin 25°× sin 25°) + (cos 25° × cos 25°)

= sin² 25° + cos² 25°

= 1

3. (i) sin 62° – cos 28°

(ii) cosec 35° – sec 55°.

Solution

(i) sin 62° – cos 28°

It can be written as

= sin (90° – 28°) – cos 28°

So we get

= cos 28° – cos 28°

= 0

(ii) cosec 35° – sec 55°

It can be written as

= cosec 35° – sec (90° – 35°)

So we get

= cosec 35° – cosec 35°

= 0

4. (i) cos² 26° + cos 64° sin 26° + tan 36°/ cot 54°

(ii) sec 17°/cosec 73° + tan 68°/cot 22° + cos² 44° + cos² 46°.

Solution

(i) cos² 26° + cos 64° sin 26° + tan 36°/ cot 54°

It can be written as

= cos² 26° + cos (90° – 26°) sin 26° + tan 36°/cot (90° – 36°)

We know that

cos (90° – θ) = sin θ

cot (90° – θ) = tan θ

sin² θ + cos² θ = 1

So we get

= cos² 26° + sin² 26° + tan 36°/tan 36°

= 1 + 1

= 2

(ii) sec 17°/cosec 73° + tan 68°/cot 22° + cos² 44° + cos² 46°

It can be written as

= sec (90° – 73°)/cosec 73° + tan (90° – 22°)/cot 22° + cos² (90° – 46°) + cos² 46°

By further calculation

= cosec 73°/cosec 73° + cot 22°/cot 22° + sin² 46° + cos² 46°

We know that sin² θ + cos² θ = 1

So we get

= 1 + 1 + 1

= 3

5. (i) cos 65°/sin 25° + cos 32°/sin 58° – sin 28° sec 62° + cosec² 30°

(ii) sec 29°/ cosec 61° + 2 cot 8° cot 17° cot 45° cot 73° cot 82° – 3 (sin² 38° + sin² 52°).

Solution

(i) cos 65°/sin 25° + cos 32°/sin 58° – sin 28° sec 62° + cosec² 30°

It can be written as

= cos 65°/sin (90° – 65°) + cos 32°/sin (90°–32°) – sin 28° sec (90°–28°) + cosec² 30°

By further calculation

= cos 65°/cos 65° + cos 32°/cos 32° – sin 28° cosec 28° + cosec² 30°

We know that cosec 30° = 2

= 1 + 1 – 1 + 4

= 5

(ii) sec 29°/cosec 61° + 2 cot8° cot17° cot45° cot73° cot82° – 3 (sin² 38° + sin² 52°)

It can be written as

= sec 29°/cosec (90° – 29°) + 2 cot8° cot17° cot45° cot (90°–17°) cot (90°–8°) – 3 [sin²38° + sin² (90°–38°)]

By further calculation

= sec 29°/sec 29° + (2 cot8° cot17° × 1 × tan17° tan8°) – 3 (sin² 38° + cos² 38°)

So we get

= 1 + (2 cot 8° tan 8° cot 17° tan 17° × 1) – (3 × 1)

We know that

cosec (90° – θ) = sec θ

⇒ cot (90° – 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 0° to 45°:

(i) tan 81° + cos 72°

(ii) cot 49° + cosec 87°.

Solution

(i) tan 81° + cos 72°

It can be written as

= tan (90° – 9°) + cos (90° – 18°)

So we get

= cot 9° + sin 18°

(ii) cot 49° + cosec 87°

It can be written as

= cot (90° – 41°) + cosec (90° – 3°)

So we get

= tan 41° + sec 3°

 

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

7. (i) sin² 28° – cos² 62° = 0

(ii) cos² 25° + cos² 65° = 1

(iii) cosec² 67° – tan² 23° = 1

(iv) sec² 22° – cot² 68° = 1.

Solution

(i) sin² 28° – cos² 62° = 0

Consider

LHS = sin² 28° – cos² 62°

It can be written as

= sin² 28° – cos² (90° – 28°)

So we get

= sin² 28° – sin² 28°

= 0

= RHS

(ii) cos² 25° + cos² 65° = 1

Consider

LHS = cos² 25° + cos² 65°

It can be written as

= cos² 25° + cos² (90° – 25°)

We know that sin² θ + cos² θ = 1

So we get

= cos² 25° + sin² 25°

= 1

(iii) cosec² 67° – tan² 23° = 1

Consider

LHS = cosec² 67° – tan² 23°

It can be written as

= cosec² 67° – tan² (90° – 67°)

We know that cosec² θ – cot² θ = 1

So we get

= cosec² 67° – cot² 67°

= 1

(iv) sec² 22° – cot² 68° = 1

Consider

LHS = sec² 22° – cot² 68°

It can be written as

= sec² 22° – cot² (90° – 22°)

We know that sec² θ – tan² θ = 1

So we get

= sec² 22° – tan² 22°

= 1

8. (i) sin 63° cos 27° + cos 63° sin 27° = 1

(ii) sec 31° sin 59° + cos 31° cosec 59° = 2.

Solution

(i) sin 63° cos 27° + cos 63° sin 27° = 1

Consider

LHS = sin 63° cos 27° + cos 63° sin 27°

It can be written as

= sin 63° cos (90° – 63°) + cos 63° sin (90° – 63°)

= sin 63° sin 63° + cos 63° cos 63°

We know that sin² θ + cos² θ = 1

So we get

= sin² 63° + cos² 63°

= 1

(ii) sec 31° sin 59° + cos 31° cosec 59° = 2

Consider

LHS = sec 31° sin 59° + cos 31° cosec 59°

It can be written as

= 1/cos 31° × sin 59° + (cos 31° × 1/sin 59°)

By further calculation

= sin 59°/cos (90° – 59°) + cos 31°/sin (90° – 31°)

So we get

= sin 59°/sin 59° + cos 31°/cos 31°

= 1 + 1

= 2

= RHS

 

9. (i) sec 70° sin 20° – cos 20° cosec 70° = 0

(ii) sin² 20° + sin² 70° – tan² 45° = 0.

Solution

(i) sec 70° sin 20° – cos 20° cosec 70° = 0

Consider

LHS = sec 70° sin 20° – cos 20° cosec 70°

By further simplification

= sin 20°/cos 70° – cos 20°/sin 70°

It can be written as

= sin 20°/cos (90° – 20°) – cos 20°/sin (90° – 20°)

So we get

= sin 20°/sin 20° – cos 20°/cos 20°

= 1 – 1

= 0

= RHS

(ii) sin² 20° + sin² 70° – tan² 45° = 0

Consider

LHS = sin² 20° + sin² 70° – tan² 45°

It can be written as

= sin² 20° + sin² (90° – 20°) – tan² 45°

By further calculation

= sin² 20° + cos² 20° – tan² 45°

We know that sin² θ + cos² θ = 1 and tan 45° = 1

So we get

= 1 – 1

= 0

= RHS

10. (i) cot 54°/tan 36° + tan 20°/cot 70° – 2 = 0

(ii) sin 50°/cos 40° + cosec 40°/sec 50° – 4 cos 50° cosec 40° + 2 = 0.

Solution

(i) cot 54°/tan 36° + tan 20°/cot 70° – 2 = 0

Consider

LHS = cot 54°/tan 36° + tan 20°/cot 70° – 2

It can be written as

= cot 54°/tan (90° – 54°) + tan 20°/cot (90° – 20°) – 2

By further calculation

= cot 54°/ cot 54° + tan 20°/ tan 20° – 2

So we get

= 1 + 1 – 2

= 0

= RHS

(ii) sin 50°/cos 40° + cosec 40°/sec 50° – 4 cos 50° cosec 40° + 2 = 0

Consider

LHS = sin 50°/cos 40° + cosec 40°/sec 50° – 4 cos 50° cosec 40° + 2

It can be written as

= sin 50°/cos (90° – 50°) + cosec 40°/sec (90° – 40°) – 4 cos 50° cosec (90° – 50°) + 2

By further calculation

= sin 50°/sin 50° + cosec 40°/cosec 40° – cos 50° sec 50° + 2

So we get

= 1 + 1 – (4 cos50°/cos 50°) + 2

= 1 + 1 – 4 + 2

= 4 – 4

= 0

= RHS

 

11. (i) cos 70°/sin 20° + cos 59°/sin 31° – 8 sin² 30° = 0

(ii) cos 80°/sin 10° + cos 59° cosec 31° = 2.

Solution

(i) cos 70°/sin 20° + cos 59°/sin 31° – 8 sin² 30° = 0

Consider

LHS = cos 70°/sin 20° + cos 59°/sin 31° – 8 sin² 30°

It can be written as

= cos 70°/sin (90° – 70°) + cos 59°/sin (90° – 59°) – 8 sin² 30°

We know that sin 30° = ½

= cos 70°/cos 70° + cos 59°/ cos 59° – 8 (1/2)²

By further calculation

= 1 + 1 – (8 × ¼)

So we get

= 2 – 2

= 0

= RHS

(ii) cos 80°/sin 10° + cos 59° cosec 31° = 2

Consider

LHS = cos 80°/sin 10° + cos 59° cosec 31°

It can be written as

= cos 80°/sin (90° – 80°) + cos 59°/ sin 31°

By further simplification

= cos 80°/ cos 80° + cos 59°/ sin (90° – 59°)

So we get

= 1 + cos 59°/ cos 59°

= 1 + 1

= 2

= RHS

12. Without using trigonometrical tables, evaluate:

(iii) sin² 34° + sin² 56° + 2 tan 18° tan 72° – cot² 30°.

Solution

So we get

= 2 + 1 – 3

= 0

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

= 1/1

= 1

(iii) sin² 34° + sin² 56° + 2 tan 18° tan 72° – cot² 30°

It can be written as

= sin² 34° + [sin (90° – 34°)]² + 2 tan 18° tan (90° – 18°) – cot² 30°

By further simplification

= sin² 34° + cos² 34° + 2 tan 18° cot 18° – (√3)²

So we get

= 1 + (2 tan 18°× 1/tan 18°) – 3

= 1 + 2 – 3

= 0

13. Prove the following:

Solution

We know that

LHS =

So we get

= cos θ/cos θ + sin θ/sin θ

= 1 + 1

= 2

= RHS

(ii) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1

Consider

LHS = cos θ sin (90° – θ) + sin θ cos (90° – θ)

It can be written as

= cos θ. cos θ + sin θ. sin θ

So we get

= cos² θ + sin² θ

= 1

= RHS

Consider

LHS =

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:

Solution

Consider

LHS =

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

Consider

LHS =

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:

Solution

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

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

Solution

Consider

We know that cos (90° – θ) = sin θ, tan (90° – 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 – 6°), where 3A and A – 6° are acute angles

(ii) tan 2A = cot (A – 18°), where 2A and A – 18° are acute angles

(iii) If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A.

Solution

(i) sin 3A = cos (A – 6°), where 3A and A – 6° are acute angles

It is given that

sin 3A = cos (A – 6°)

We know that cos (90° – θ) = sin θ

cos (90° – 3A) = cos (A – 6°)

By comparing both

90° – 3A = A – 6°

By further calculation

90° + 6° = A + 3A

⇒ 96° = 4A

So we get

A = 96°/4 = 24°

Hence, the value of A is 24°.

(ii) tan 2A = cot (A – 18°)

We know that cot (90° – θ) = tan θ

cot (90° – 2A) = cot (A – 18°)

By comparing both

90° – 2A = A – 18°

By further calculation

90° + 18° = A + 2A

So we get

3A = 108°

⇒ A = 108°/3 = 36°

Hence, the value of A is 36°.

(iii) sec 2A = cosec (A – 27°)

We know that cosec (90° – θ) = sec θ

cosec (90° – 2A) = cos (A – 27°)

By comparing both

90° – 2A = A – 27°

By further calculation

90° + 27° = A + 2A

So we get

3A = 117°

⇒ A = 117°/3 = 39°

Hence, the value of A is 39°.

18. Find the value of θ (0° < θ < 90°) if:

(i) cos 63° sec (90° – θ) = 1

(ii) tan 35° cot (90° – θ) = 1.

Solution

(i) cos 63° sec (90° – θ) = 1

It can be written as

cos 63° = 1/sec(90° – θ)

We know that 1/sec θ = cos θ

cos 63° = cos (90° – θ)

By comparing both

90° – θ = 63°

By further calculation

θ = 90° – 63° = 27°

(ii) tan 35° cot (90° – θ) = 1

It can be written as

tan 35° = 1/cot (90° – θ)

We know that 1/cot θ = cos θ

tan 35° = tan (90° – θ)

By comparing both

35° = 90° – θ

By further calculation

θ = 90° – 35° = 55°

 

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 = 180°

Dividing both sides by 2

(∠A + ∠B + ∠C)/2 = 180°/2

⇒ A/2 + B/2 + C/2 = 90°

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

We can write it as

(A + B)/2 = 90° – C/2

We know that

cos (90° – C/2) = sin C/2

Here, cos (90° – θ) = sin θ

sin C/2 = sin C/2

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

We know that (A + C)/2 = 90° – B/2

= tan (90° – B/2)

So we get

= cot B/2

= RHS

Chapter Test

1. Find the values of:
(i) sin² 60° – cos² 45° + 3tan² 30°
(ii)
(iii) sec 30° tan 60° + sin 45° cosec 45° + cos 30° cot 60°

Solution

(i) sin² 60° – cos² 45° + 3tan² 30°

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

Hence,

= 1

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

= 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°

Now,

R.H.S. = cos 2A = cos 2(30^o) = 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,

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°

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 60^o

⇒ 2θ = 60^o

⇒ θ = 30^o

Now,

sin θ + √3 cos θ – 2 tan² θ

= sin 30^o + √3 cos 30^o – 2 tan² 30^o

= 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 + A - B = 60° + 45°

⇒ 2A = 105^o

⇒ A = 52½^o

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

52½^o + B = 60°

⇒ B = 60° – 52½^o = 7½^o

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

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

(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

= (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°

= 5 – 3

= 2

9. Prove that:

Solution

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°

⇒  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.

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.