Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1

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

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°)

(v) -230° = 360° – 230° = 130° = (90° + 40°) II quadrant

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°

Question 3.

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

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

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

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

Question 11.
If cosec θ – sin θ = a³ and sec θ – cos θ = b³, then prove that a²b² (a² + b²) = 1.
Solution:

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

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

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