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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} = c^{2}

(i.e) a^{2} cos^{2} θ + b^{2} sin^{2} θ – 2ab sin θ cos θ = c^{2}

(i.e) a^{2} (1 – sin^{2} θ) + b^{2} (1 – cos^{2} θ) – 2ab sin θ cos θ = c^{2}

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

a^{2} + b^{2} – c^{2} = a^{2} sin^{2} θ + b^{2} cos^{2} 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 sec^{2} θ – tan^{2} 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 (m^{2} – n^{2})^{2} = mn.

Solution:

cot θ (1 + sin θ) = 4m

Question 11.

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

Solution:

(sec θ + cos θ) (sec θ – cos θ) = sec^{2} θ – cos^{2} θ

= (1 + tan^{2} θ ) – (1 – sin^{2} θ)

= tan^{2} θ + sin^{2} θ = RHS

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

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)