Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.1 Text Book Back Questions and Answers, Notes.

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°
(v) – 230°
Answer:
(i) 25°

25° First quadrant

(ii) 825°

825° = 9 × 90° + 15°
825° = 2 × 360° + 105°
∴ 825° lies in the second quadrant.

iii) -55°

-55° lies in the fourth quadrant

iv) 328°

328° = 270° + 58° lies in the fourth quadrant.

v) -230°

– 230° = – 180° + (- 50°) lies in the second quadrant.

Question 2.
For each given angle, find a co-terminal angle with a measure of 9 such that 0 ≤ θ < 360°.
(i) 395°
(ii) 525°
(iii) 1150°
(iv) – 270°
(v) – 450°
Answer:
(i) 395°
395° = 360° + 35°
395° – 35° = 360°
∴ Coterminal angle for 395° is 35°

(ii) 525°
525° = 360° + 165°
360° – 165° = 360°
∴Coterminal angle for 525° is 165°

(iii) 1150°
1150° = 360° + 360° + 360° + 70°
1150° = 3 × 360° + 70°
1150° – 70° = 3 × 360°
∴ Coterminal angle for 1150° is 70°.

(iv) – 270°
– 270° = 360° + 90°
– 270° – 90° = 360°
∴ Coterminal angle for -270° is 90°

(v) – 450°
– 450° = – 720° + 270°
– 450° – 270° = – 2 × 360°
∴ Coterminal angle for – 450° is 270°

Question 3.
If a cos θ – b sin θ = c , show that a sin θ + b cos θ = ± \(\sqrt{\mathbf{a}^{2}+\mathbf{b}^{2}-\mathbf{c}^{2}}\)
Answer:
a cos θ – b sin θ = c
(a cos θ – b sin θ)² + (a sin θ + b cos θ)² = a² cos² θ – 2 ab sin θ cos θ + b² sin²θ + a² sin² θ + b² cos² θ + 2 ab sin θ cos θ
c² + (a sin 0 + b cos θ )² = a² cos² θ + a² sin² θ + b² sin² θ + b²cos²θ
= a² (cos²θ + sin²θ) + b²(sin²θ + cos²θ)
c² + (a sin θ + b cos θ )2 = a² + b²
(a sin θ + b cos θ)² = a² + b² – c²
a sin θ + b cos θ = ± \(\sqrt{\mathbf{a}^{2}+\mathbf{b}^{2}-\mathbf{c}^{2}}\)

Question 4.
If sin θ + cos θ = m , show that cos⁶ θ + sin⁶ θ = \(\frac{4-3\left(m^{2}-1\right)^{2}}{4}\) where m² ≤ 2.
Answer:
sin θ + cos θ = m
(sin θ + cos θ)² = m²

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.
If tan² θ = 1 – k², show that sec θ + tan³ θ cosec θ = ( 2 – k²)^3/2. Also, find the values of k for which this result holds.
Answer:
tan² θ = 1 – k²
1 + tan² θ = 1 + 1 – k²
sec²θ = (2 – k²)
sec²θ = (2 – k²)^1/2


tan² θ = 1 – k²
When θ = \(\frac{\pi}{2}\), tan \(\frac{\pi}{2}\) = ∞, not defined 2
When θ = 0, tan² 0 = 1 – k²
1 – k² = 0 ⇒ k² = 1 ⇒ k = ± 1
When θ = 45°, tan² 45° = 1 – k²
1 – k² = 1 ⇒ – k² = 0 ⇒ k = 0
When θ > 45°, say θ = 60°
tan² 60° = 1 – k² = (√3)² = 1 – k²
3 = 1 – k² ⇒ k² = 1 – 3 = – 2
∴ θ > 45°, k² is negative ⇒ k is imaginary
∴ k lies between -1 and 1 ⇒ k ∈ [-1 , 1]

Question 9.
If sec θ + tan θ = p, obtain the values of sec θ, tan θ and sin θ in terms of p.
Answer:
Given sec θ + tan θ = p
We have sec² θ – tan² θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1
p (sec θ – tan θ) = 1
sec θ – tan θ = \(\frac{1}{p}\)
(sec θ – tan θ) + (sec θ – tan θ) = p + \(\frac{1}{p}\)

Question 10.
If cot θ(1 + sin θ) = 4m and cot θ (1 – sin θ) = 4n then prove that (m² – n²)² = mn.
Answer:

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

Question 12.
Eliminate θ from the equations a sec θ – c tan θ = b, b sec θ + d tan θ = c.
Answer: