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 }= a^{2} cos^{2} θ – 2 ab sin θ cos θ + b^{2} sin^{2}θ + a^{2} sin^{2} θ + b^{2} cos^{2} θ + 2 ab sin θ cos θ

c^{2} + (a sin 0 + b cos θ )^{2} = a^{2} cos^{2} θ + a^{2} sin^{2} θ + b^{2} sin^{2} θ + b^{2}cos^{2}θ

= a^{2} (cos^{2}θ + sin^{2}θ) + b^{2}(sin^{2}θ + cos^{2}θ)

c^{2} + (a sin θ + b cos θ )2 = a^{2} + b^{2}

(a sin θ + b cos θ)^{2} = a^{2} + b^{2} – c^{2}

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 cos^{6} θ + sin^{6} θ = \(\frac{4-3\left(m^{2}-1\right)^{2}}{4}\) where m^{2} ≤ 2.

Answer:

sin θ + cos θ = m

(sin θ + cos θ)^{2} = m^{2}

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

If tan^{2} θ = 1 – k^{2}, show that sec θ + tan^{3} θ cosec θ = ( 2 – k^{2})^{3/2}. Also, find the values of k for which this result holds.

Answer:

tan^{2} θ = 1 – k^{2}

1 + tan^{2} θ = 1 + 1 – k^{2}

sec^{2}θ = (2 – k^{2})

sec^{2}θ = (2 – k^{2})^{1/2}

tan^{2} θ = 1 – k^{2}

When θ = \(\frac{\pi}{2}\), tan \(\frac{\pi}{2}\) = ∞, not defined 2

When θ = 0, tan^{2} 0 = 1 – k^{2}

1 – k^{2} = 0 ⇒ k^{2} = 1 ⇒ k = ± 1

When θ = 45°, tan^{2} 45° = 1 – k^{2}

1 – k^{2} = 1 ⇒ – k^{2} = 0 ⇒ k = 0

When θ > 45°, say θ = 60°

tan^{2} 60° = 1 – k^{2} = (√3)^{2} = 1 – k^{2}

3 = 1 – k^{2} ⇒ k^{2} = 1 – 3 = – 2

∴ θ > 45°, k^{2} 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 sec^{2} θ – tan^{2} 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 (m^{2} – n^{2})^{2} = mn.

Answer:

Question 11.

If cosec θ – sin θ = a^{3}, sec θ – cos θ = b^{3} then prove that a^{2}b^{2}(a^{2} + b^{2}) = 1.

Answer:

Question 12.

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

Answer: