# Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3

Question 1.
Find the values of
(i) sin(480°)
(ii) sin(-1110°)
(iii) cos(300°)
(iv) tan(1050°)
(v) cot(660°)
(vi) tan $$\left(\frac{19 \pi}{3}\right)$$
(vii) sin $$\left(-\frac{11 \pi}{3}\right)$$
Solution:
(i) sin(480°) = sin(360° + 120°) = sin 120°
= sin(90° + 30°) = cos 30° = $$\sqrt{3}$$/2

(ii) sin(-1110°) = -sin(1110°)
= – sin (360° × 3 + 30°)
= -sin 30° = -1/2

(iii) cos(300°) = cos(270° + 30°) = sin 30° = 1/2

(iv) tan(1050°) = tan [3(360°) – 30°]

(v) cot(660°) = cot (360° × 2 – 60°)

Question 2.
$$\left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right)$$ is a point on the terminal side of an angle θ is standard position. Determine the trigonometric function values of angle θ .
Solution:

Question 3.
Find the values of other five trigonometric functions for the following:

(i) cos θ = $$-\frac{1}{2}$$; θ lies in the III quadrant.
Solution:
Taking the Numerical values

(ii) cos θ = $$\frac{2}{3}$$ ; θ lies in the I quadrant
Solution:

(iii) sin θ = –$$\frac{2}{3}$$ ; θ lies in the IV quadrant
Solution:

(iv) tan θ = -2; θ lies in the II quadrant
Solution:

(v) sec θ = $$\frac{13}{5}$$ ; θ lies in the IV quadrant
Solution:

Question 4.

Solution:
cot(180° + θ) = cot θ
sin (90° – θ) = cos θ
cos(-θ) = cos θ
sin (270 + θ) = – cos θ
tan(-θ) = -tan θ
cosec (360° + θ) = cosec θ

Question 5.
Find all the angles between 0° and 360° which satisfy the equation sin2 θ = $$\frac{3}{4}$$
Solution:

Question 6.

Solution:
LHS = sin2 10° + sin2 20° + sin2 70° + sin2 80°
= sin2 10° + sin2 (90° – 10°) + sin2 20° + sin2(90° – 20°)
= sin2 10° + (cos 10°)2 + sin2 20° + (cos 20°)2
= (sin2 10+ cos2 10) + sin2 20° + cos2 20°
= 1 + 1 = 2 = RHS

### Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3 Additional Questions Solved

Question 1.
Prove that: sin 600°. tan (-690°) + sec 840°. cot (-945°) = $$\frac{3}{2}$$
Solution:

Question 2.
Prove that sin (270° – θ) sin (90° – θ) – cos (270° – θ) cos (90° + θ) + 1 = 0
Solution:
LHS = sin (270° – θ) sin (90° – θ) – cos (270° – θ) cos (90° + θ) + 1
Now, sin (270° – θ) = sin {180°+ (90°- θ)}
= – cos (90° – θ) = – sin θ
LHS = – cos θ . cos θ – (- sin θ) (- sin θ) + 1
= – cos2 θ – sin2 θ + 1
= – (cos2 θ + sin2 θ) + 1 = -1 + 1 = 0 = RHS

Question 3.
Prove that cos 24° + cos 55° + cos 125° + cos 204° + cos 300° = $$\frac{1}{2}$$
Solution:
cos 204° = cos (180°+ 24°) = – cos 24°
cos 125° = cos (180° – 55°) = – cos 55°
LHS = cos 24° + cos 55° + (- cos 55°) + (- cos 24°) + cos 300°
= cos 24° + cos 55° – cos 55° – cos 24° + cos 300°

Question 4.

Solution:

Question 5.

Solution:

LHS = [(1 +cot α) + cosec α][(1 + cot α) – cosec α]
= (1 + cot α)2 – cosec2 α
= 1 + cot2α + 2 cot α – cosec2 α
[∵ 1 + cot2α = cosec2α]
= cosec2α + 2 cot α – cosec2α
= 2 cot α = RHS

Question 6.

Solution:

= sec (450° – θ)
= sec[360° + (90° – θ)]
sec (90° – θ) = cosec θ

Question 7.

Solution:
cos (90° + θ) = – sin θ
sec (- θ) = sec θ
tan (180° – θ) = – tan θ
sec (360° – θ) = sec θ
sin (180° + θ) = – sin θ
cot (90° + θ) = – tan θ

Question 8.
Find x from the equation cosec (90° + A) + x cos A cot (90° + A) = sin (90° + A).
Solution: