You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1

11th Maths Exercise 3.1 Answers Question 1.
Identify the quadrant in which an angle of each given measure lies
(i) 25°
(ii) 825°
(iii) -55°
(iv) 328°
Solution:
(i) 25° = I quadrant
(ii) 825° = 105° (90° + 15°) = II quadrant
(iii) -55° = IV quadrant
(iv) 328° = IV quadrant (270° + 58°)
11th Maths Exercise 3.1 Answers Solutions Chapter 3 Trigonometry Samacheer Kalvi
(v) -230° = 360° – 230° = 130° = (90° + 40°) II quadrant

11th Maths Exercise 3.1 Question 2.
For each given angle, find a coterminal angle with measure of θ such that θ° < θ < 360°
(i) 395°
(ii) 525°
(iii) 1150°
(iv) -270°
(v) -450°
Solution:
(i) 395° = 360° + 35°
∴ coterminal angle = 35°

(ii) 525° – 360°= 165°
coterminal angle = 165°

(iii) 1150° = 360 × 3 + 70° = 70°
coterminal angle = 70°

(iv) -270° = coterminal angle =+90° {270° + 90° = 360°}

(v) -450° = -360° – 90° = -90°
∴ coterminal angle = 360° – 90° = 270°

Exercise 3.1 Class 11 Maths Solutions State Board Question 3.
11th Maths Exercise 3.1 Solutions Chapter 3 Trigonometry Samacheer Kalvi
Solution:
a cos θ – b sin θ = c
⇒ (a cos θ – b sin θ)2 = c2
(i.e) a2 cos2 θ + b2 sin2 θ – 2ab sin θ cos θ = c2
(i.e) a2 (1 – sin2 θ) + b2 (1 – cos2 θ) – 2ab sin θ cos θ = c2
a2 – a2 sin2 θ + b2 – b2 cos2 θ – 2ab sin θ cos θ = c2
a2 + b2 – c2 = a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 20

Exercise 3.1 Class 11 Maths State Board Question 4.
Exercise 3.1 Class 11 Maths Solutions State Board Chapter 3 Trigonometry Samacheer Kalvi
Solution:
Exercise 3.1 Class 11 Maths State Board Chapter 3 Trigonometry Ex 3.1

Samacheer 11th Maths Solution Question 5.
Samacheer 11th Maths Solution Chapter 3 Trigonometry Ex 3.1
Solution:
Samacheer Kalvi 11 Maths Solutions Chapter 3 Trigonometry Ex 3.1
Samacheer Kalvi 11th Maths Guide Solutions Chapter 3 Trigonometry Ex 3.1

Samacheer Kalvi 11 Maths Solutions Question 6.
Samacheer Kalvi 11 Maths Solutions Chapter 3 Trigonometry Ex 3.1
Solution:
Samacheer Kalvi 11th Maths Example Sums Chapter 3 Trigonometry Ex 3.1

Samacheer Kalvi 11th Maths Guide Question 7.
Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Ex 3.1
Solution:
Samacheer Kalvi Class 11 Maths Solutions Chapter 3 Trigonometry Ex 3.1
Samacheer Class 11 Maths Solutions Chapter 3 Trigonometry Ex 3.1

Samacheer Kalvi 11 Maths Question 8.
11th Maths 3rd Chapter Solutions Trigonometry Ex 3.1 Samacheer Kalvi
Solution:
11th Maths 3rd Chapter Solutions Trigonometry Ex 3.1 Samacheer Kalvi

Samacheer Kalvi 11th Maths Example Sums Question 9.
If sec θ + tan θ = p, obtain the values of sec θ, tan θ and sin θ in terms of p.
Solution:
Given, sec θ + tan θ = p
we know sec2 θ – tan2 θ = 1
(i.e) (sec θ + tan θ) (sec θ – tan θ) = 1
Class 11 Maths Solutions Samacheer Kalvi Chapter 3 Trigonometry Ex 3.1

Samacheer Kalvi Class 11 Maths Solutions Question 10.
If cot θ (1 + sin θ) = 4m and cot θ (1 – sin θ) = 4n, then prove that (m2 – n2)2 = mn.
Solution:
cot θ (1 + sin θ) = 4m
11th Samacheer Maths Solutions Chapter 3 Trigonometry Ex 3.1
Samacheer Kalvi Guru 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1
Samacheer Kalvi 11th Maths Chapter 3 Trigonometry Ex 3.1

Samacheer Kalvi Class 11 Maths Question 11.
If cosec θ – sin θ = a3 and sec θ – cos θ = b3, then prove that a2b2 (a2 + b2) = 1.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1
Samacheer Kalvi 11th Maths Solution Chapter 3 Trigonometry Ex 3.1
11 Samacheer Maths Solutions Chapter 3 Trigonometry Ex 3.1

Samacheer Class 11 Maths Solutions Question 12.
Eliminate θ from the equations a sec θ – c tan θ = b and b sec θ + d tan θ = c.
Solution:
Taking sec θ = X and tan θ = Y we get the equations as
11th Maths Solutions Samacheer Kalvi Chapter 3 Trigonometry Ex 3.1
Samacheer Kalvi 11th Maths Book Back Answers Chapter 3 Trigonometry Ex 3.1

Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 Additional Equations

11th Maths 3rd Chapter Solutions Question 1.
Prove that (sec θ + cos θ) (sec θ – cos θ) = tan2 θ + sin2 θ
Solution:
(sec θ + cos θ) (sec θ – cos θ) = sec2 θ – cos2 θ
= (1 + tan2 θ ) – (1 – sin2 θ)
= tan2 θ + sin2 θ = RHS

Class 11 Maths Solutions Samacheer Kalvi Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 50
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 51
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 52

11th Samacheer Maths Solutions Question 3.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 53
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 54

Samacheer Kalvi Guru 11th Maths Question 4.
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 55
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 56

Samacheer Kalvi 11th Maths Question 5.
Prove that (1 + tan A + sec A) (1 + cot A – cosec A) = 2
Solution:
LHS = (1 + tan A + sec A) (1 + cot A – cosec A)
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 57
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1 58

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