Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.10 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.10

Question 1.
Determine whether the following measurements produce one triangle, two triangles or no triangle. ∠B = 88°, a = 23 , b = 2. Solve if solution exists.
Answer:
Using sine formula

= 23 × 0.999
= 22.99
which is not possible
∴ Solution of the given triangle does not exsit.

Question 2.
If the sides of a ∆ ABC are a = 4, b = 6 and C = 8, then show that 4 cos B + 3 cos C = 2.
Answer:
In ∆ ABC Given that a = 4, b = 6, c = 8
Using cosine formula

Question 3.
In a ∆ ABC, if a = √3 – 1, b = √3 + 1 and C = 60° find the other side and other two angles.
Answer:

In a ∆ ABC, Given
a = √3 – 1, b = √3 + 1
C = 60°
Using cosine formula a
C² = a² + b² – 2 ab cos C
= (√3 – 1)² + (√3 + 1)² – 2(√3 – 1) × (√3 + 1) cos 6o°
= 3 – 2√3 + 1 + 3 + 2√3 + 1 – 2 (3 – 1) × \(\frac { 1 }{ 2 }\)
c² = 8 – 2 = 6 ⇒ c = √6
Using sine formula

sin (45° – 30°) = sin 45° . cos 30° – cos 45° sin 30°

From equations (1) and (2), we have
sin A = sin 15° ⇒ A = 15°
In ∆ ABC, we have A + B + C = 180°
15° + B + 60° = 180°
B = 180°- 75°
B = 105°
∴ The required sides and angles are
c = √6, A = 15°, B = 105°

Question 4.
In any ∆ ABC, prove that the area

Answer:
Area of ∆ ABC is ∆ = \(\frac { 1 }{ 2 }\) bc = sin A
Using cosine formula

Question 5.
In a ∆ABC, if a = 12 cm, b = 8 cm and C = 30°, then show that its area is 24 sq.cm.
Answer:

In ∆ ABC Given
a = 12 cm ,
b = 8 cm,
C = 30°

Question 6.
In a ∆ABC, if a = 18 cm, b = 24 cm and c = 30 cm, then show that its area is 216 sq.cm.
Answer:
In a ∆ ABC, Given a = 18 cm, b = 24cm and c = 30 cm

Area of the triangle ABC

Question 7.
Two soldiers A and B in two different underground bunkers on a straighi road, spot an intruder at the top of a hill. The angle of elevation of the intruder from A and B to the ground level in the eastern direction are 30° and 45° respectively. If A and B stand 5km apart, find the distance of the intruder from B.
Answer:
Let A and B be the two positions of the soldiers.
AC – direction of the intruder seen from A.
BC – the direction of the intruder seen from B.
∠ BAC = 30° angle of elevation of the intruder from A.
∠ PBC = 45° angle of elevation of the intruder from B.

Distance between A and B = 5k.m.
In ∆ ABC, ∠ ABC = 180° – 45° = 135°
∠ BCA = 180° – ( 135° + 30°)
= 180° – 165° = 15°
Using sine formula

Question 8.
A researcher wants to determine the width of a pond from east to west, which cannot be done by actual measurement. From a point P, he finds the distance to the eastern-most point of the pond to be 8 km, while the distance to the westernmost point from P to be 6 km. If the angle between the two lines of sight is 60°, find the width of the pond.
Answer:

A – be the easternmost point on the pond and
B – be the westernmost point on the pond.
AB – Width of the pond
P – Point of observation.
The distance of A from P = 8 km
Distance of B from P = 6km
Angle between the directions PA and PB
∠APB = 60°
In ∆ PAB, using cosine formula
AB² = PA² + PB² – 2PA . PB . cos ∠APB
AB² = 8² + 6² – 2 × 8 × 6 . cos 60°
= 64 + 36 – 96 × \(\frac { 1 }{ 2 }\)
= 100 – 48 = 52
AB = \(\sqrt{52}\) = \(\sqrt{4 \times 13}\)
AB = 2\(\sqrt{13}\) k.m.
Width of the pond = 2\(\sqrt{13}\) k.m

Question 9.
Two Navy helicopters A and B are flying over the Bay of Bengal at saine altitude from sea level to search a missing boat. Pilots of both the helicopters sight the boat at the same time while they are apart 10km from each other. If the distance of the boat from A is 6 km and if the line segment AB subtends 60° at the boat, find the distance of the boat from B.
Answer:
A , B are the positions of the helicopter above the sea level.

Distance between A and B = 10 km
C – Position of the boat on the surface of sea.
AC, BC are the directions of the boat as seen from A and B respectively.
Distance of the boat C from A = 6 k.m
∠ ACB = 60°
Using cosine formula
AB² = BC² + AC² – 2 BC . AC cos ∠ACB
c² = a² + b² – 2 ab cos C
10² = a² + 6² – 2a × 6 cos 60°
100 = a² + 36 – 12a\(\left(\frac{1}{2}\right)\)
0 = a² + 36 – 6a – 100
a² – 6a – 64 = 0

Question 10.
A straight tunnel is to be made through a mountain. A surveyor observes the two extremities A and B of the tunnel to be built from a point P in front of the mountain. If AP = 3 km, BP = 5 km, and ∠APB = 120°, then find the length of the tunnel to be built.
Answer:

p² = a² + b² – 2ab cos P
p² = 9 + 25 – 30 Cos 120°
p² = 9 + 25 – 30 (-1/2) = 34 + 15 = 49
⇒ p = \(\sqrt{49}\) = 7 km

Question 11.
A farmer wants to purchase a triangular-shaped land with sides 120 feet and 60 feet and the angle included between these two sides is 60°. If the land costs Rs.500 per square feet, find the amount he needed to purchase the land. Also, find the perimeter of the land.
Answer:

Let ∆ ABC be the shape of the land.
Given AB = 120 ft, AC = 60ft
∠ BAC = 60°
Using cosine formula in ∆ ABC
BC² = AB² + AC² – 2AB . AC cos ¿BAC
BC² = 120² + 60² – 2 × 120 × 60 cos (60°)
= 14400 + 3600 – 14400 × \(\frac { 1 }{ 2 }\)
= 18000 – 7200
BC² = 10800 = 100 × 2 × 2 × 3 × 3 × 3
BC² = 10² × 2² × 3² × 3
BC = \(\sqrt{10^{2} \times 2^{2} \times 3^{2} \times 3}\)
BC = 10 × 2 × 3√3
BC = 60√3 k.m.
Perimeter of the Land = AB + BC + AC
= 120 + 60√3 + 60
= 180 + 60√3
= 60 (3 + √3) feet.
Area of ∆ ABC = \(\frac { 1 }{ 2 }\) × AB × AC × sin ∠ BAC
= \(\frac { 1 }{ 2 }\) × 60 × 120 sin 60°
= 30 × 120 × \(\frac{\sqrt{3}}{2}\)
= 30 × 60 × √3
= 1800 √3 sq. feet.
Cost of 1 sq. feet Rs. 500
∴ Cost of 800 √3 sq. feet = 800 √3 × 500 = 900000√3
Total amount needed = Rs. 900000√3
Perimeter of the land = 60(3 + √3)feet.

Question 12.
A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted, the pilot measures the angle of depression to be 30°. If after 100 km, the target has an angle of depression of 45°, how far is the target from the fighter jet at that instant?
Answer:

Let A be the position of the jet fighter observing the target at an angle of depression 30°.
Also, Let B be the position of the jet 100 k.m away horizontally from A observing the target at an angle of depression 45°.
In ∆ TAB, AB = 100 k.m
∠TAB = 30°
∠ABT = 180°- 45° = 135°
∠ATB = 180° – (135°+ 300) = 180° – 165° = 15°

Question 13.
A plane is 1 km from one landmark and 2 km from another. From the plane’s point of view, the land between them subtends an angle of 45°. How far apart are the landmarks?
Answer:
A, B are the two landmarks,
C – Position of the plane.
The distance of the plane from the landmark A = 1 k.m
The distance of the plane from the landmark B = 2 k.m
∠ACB = 45°

From the ∆ ABC, using cosine formula
AB² = AC² + BC² – 2AC. BC. cos45°
= 1² + 2² – 2 × 1 × 2
AB² = 1 + 4 – 2 × √2 = 5 – 2√2
AB = \(\sqrt{5-2 \sqrt{2}}\)
Distance between the landmarks AB = \(\sqrt{5-2 \sqrt{2}}\) km.

Question 14.
A man starts his morning walk at a point A reaches two points B and C and finally back to A such that ∠A = 60° and ∠B = 45°, AC = 4km in the ∆ ABC. Find the total distance he covered during his morning walk.
Answer:

Given In ∆ABC
AC = 4 k.m
∠A = 60°,
∠B = 45°
∠C = 180° – (60° + 45°)
∴ ∠C = 180° – 105° = 75°
Using sine formula

Question 15.
Two vehicles leave the same place P at the same time moving along two different roads. One vehicle moves at an average speed of 60 km/hr and the other vehicle moves at an average speed of 80 km/hr. After half an hour the vehicle reaches destinations A and B. If AB subtends 60° at the initial point P, then find AB.
Answer:

P – Initial point.
PA – The direction of the first vehicle travels with speed km/hr.
PB – The direction of the second vehicle travels with a speed of 80km/hr.
Given in half an hour first vehicle reaches destination A.
∴ PA = \(\frac{60}{2}\) = 30 km.
Also in half an hour the second vehicle reaches the destination B.
∴ PA = \(\frac{80}{2}\) = km.
In ∆ PAB, PA = 30, PB = 40, ∠APB = 60°
Using cosine formula
AB² = PA² + PB² – 2PA PB cos ∠APB
AB = 30² + 40² – 2 × 30 × 40 cos 60°
= 900 + 1600 – 2400 × \(\frac { 1 }{ 2 }\)
= 2500 – 1200
AB² = 1300
AB = \(\sqrt{1300}\) = \(\sqrt{13 \times 100}\)
AB = 10√13 k.m.

Question 16.
Suppose that a satellite in space, an earth station, and the centre of earth all lie in the same plane. Let r be the radius of earth and R he the distance from the centre of earth to the satellite. Let d be the distance from the earth station to the satellite. Let 30° be the angle of elevation from the earth station to the satellite, If the line segment connecting the earth station and satellite subtends angle α at the centre of earth then prove that

Answer:

O – Centre of Earth,
A – Position of Earth station.
S – Position of the satellite.
Given the radius of Earth
OA = r
The angle of elevation of the satellite from the Earth station = 30°
The distance of the satellite from the Earth station AS = d
The distance of the satellite from the centre of the Earth OS = R.
Angle subtended by the line segment AS at the centre of earth ∠AOS = α
In △ AOS, OA = r, AS = d, OS = R, ∠AOS = α
Using cosine formula
AS² = OA² + OS² – 2 OA . OS cos ∠AOS
d² = r² + R² – 2(r) (R) cos α

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.9

Question 1.
In a ∆ ABC, if \(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\) prove that a², b², C² are in Arithmetic progression.
Answer:
\(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\)
sin A . sin (B – C) = sin C . sin (A – B)
sin (180° – (B + C)) . sin (B – C) = sin (180° – (A + B)) . sin (A – B)
sin (B + C) sin (B – C) = sin (A + B) sin (A – B) ——— (1)
sin(B + C) . sin(B – C) = (sin B cos C + cos B sin C) × (sin B cos C – cos B sin C)
= (sin B cos C)² – (cos B sin C)²
= sin² B cos² C – cos² B sin² C
= sin² B (1 – sin² C) – (1 – sin² B) sin² C
= sin² B – sin² B sin² C – sin² C + sin² B sin² C
sin ( B + C) . sin ( B – C) = sin² B – sin² C
Similarly,
sin (A + B ) . sin (A – B) = sin² A – sin² B
(1) ⇒ sin² B – sin² C = sin² A – sin² B
sin² B + sin² B = sin² A + sin² C
2 sin² B = sin² A + sin² C ——— (2)

Question 2.
The angles of a triangle A B C, are in Arithmetic Progression and if b : c = √3 : √2 find ∠A.
Answer:
Given that the angles A, B, C are in A. P.
∴ 2B = A + C
Also A + B + C = 180°
B + (A + C) = 180°
B + 2B = 180°
3B = 180° ⇒ B = 60°
A + C = 2B = 2 × 60° = 120°

A + 45° = 120°
A = 120° – 45° = 75°
A = 75°

Question 3.
In a ∆ ABC, if cos c = \(\frac{\sin \mathbf{A}}{2 \sin B}\) show that the triangle is isosceles.
Answer:

a² + b² – c² = a²
b² – c² = 0
b² = c² ⇒ b = c
Two sides of is ∆ ABC are equal.
∴ ∆ ABC is an isosceles triangle.

Question 4.
In a ∆ ABC, prove that

Answer:

Question 5.
In an ∆ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.
Answer:
LHS = a cos A+ 6 cos B + c cos C
Using sine formula, we get k sin A cos A + k sin B cos B + k sin C cos C k
= \(\frac{k}{2}\) [2 sin A cos A + 2 sin B cos B + 2 sin C cos C]
= \(\frac{k}{2}\) [sin 2A + sin 2B + sin 2C]
= \(\frac{k}{2}\) [2 sin (A + B) . cos (A – B) + 2 sin C . cos C]
= \(\frac{k}{2}\) [2 sin (A – B) . cos (A – B) + 2 sin C . cos C]
= \(\frac{k}{2}\) [2 sin C . cos (A – B) + 2 sin C . cos C]
= k sin C [cos(A – B) + cos C]

= k sin C [cos (A – B) – cos (A + B)]
= k sin C . 2 sin A sin B
= 2k sin A . sin B sin C
= 2a sin B sin C
= RHS

Question 6.
In a ∆ ABC, ∠A = 60°. Prove that b + c = 2a cos \(\left(\frac{\mathbf{B}-\mathbf{C}}{2}\right)\)
Answer:
Given ∠A = 60°
A + B + C = 180°
60° + B + C = 180°
B + C = 180° – 60° = 120°

Question 7.
In an ∆ ABC, prove the following,
(i) a sin \(\left(\frac{\mathbf{A}}{2}+\mathbf{B}\right)\) = (b + c) . sin \(\frac{\mathbf{A}}{2}\)
Answer:

(ii) a (cos B + cos C) = 2(b + c) sin²\(\frac{\mathbf{A}}{2}\)
Answer:

(iii)
Answer:

(iv)
Answer:

(v)
Answer:

Question 8.
In a triangle ∆ ABC, prove that
(a² – b² + c²) tan B = (a² + b² – c²) tan C
Answer:

Question 9.
An Engineer has to develop a triangular shaped park with a perimeter 120m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park.
Answer:
Let ∆ A B C be the triangular-shaped park.
a, b, c be the length of the sides.
Given perimeter of the park = 120 m
2s = a + b + c = 120m —– (1)
For a fixed perimeter 2s. the area of a triangle is maximum when a = b = c.
(1) = a + a + a = 120
3a = 120
⇒ a = 40m
Length of the sides 40 m, 40 m, 40 rn.

Question 10.
A rope of length 42m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed.
Answer:
Let a, b, c be the lengths of the sides of the triangle.
Given the perimeter of the triangle
2s = a + b + c = 42m —-—(1)
For a fixed perimeter 2 s, the area of a triangle is maximum
when a = b = c.
(1) ⇒ a + a + a = 42
3a = 42 ⇒ a = \(\frac{42}{3}\)
a = 14m
∴ a = b = c = 14m

∴ The dimensions of the triangle are 14 m, 14 m, 14 m.
Maximum area = 49√3 sq.m.

Question 11.
Derive Projection formula from
(i) Law of sines,
(ii) Law of cosines.
Answer:
To prove (a) a = b cos C + c cos B
(b) b = c cos A + a cos C
(c) c = a cos B + b cos A

(i) Using the Law of sines,

(a) b cos C + c cos B = 2 R sin B cos C + 2 R sin C cos B
= 2 R ( sin B cos C + cos B sin C)
= 2R sin (B + C)
= 2R sin (180° – A)
b cos C + c cos B = 2R sin A = a
a = b cos C + c cos B

(b) c cos A + a cos C = 2R sin C cos A + 2R sin A cos C
= 2R (sin C cos A + cos C sin A)
= 2R sin(C + A)
= 2R sin(180° – B)
= 2R sin B = b
∴ b = c cos A + a cos C

(c) a cos B + b cosA = 2R sin A cos B + 2R sin B cos A
= 2R (sin A cos B + cos A sin B)
= 2R sin (A + B)
= 2R sin (180° – C)
= 2R sin C = c
∴ c = a cos B + b cos A

(ii) Using Law of cosines.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.8 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.8

Question 1.
Find the principal solution and general solutions of the following
(i) sin θ = – \(\frac{1}{\sqrt{2}}\)
(ii) cot θ = √3
(iii) tan θ = –\(\frac{1}{\sqrt{3}}\)
Answer:
(i) sin θ = – \(\frac{1}{\sqrt{2}}\)
We know that principal of sin θ lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
sin θ = – \(\frac{1}{\sqrt{2}}\) < 0
∴ The principal value of sin θ lies in the IV quadrant.
sin θ = – \(\frac{1}{\sqrt{2}}\)
= – sin \(\left(\frac{\pi}{4}\right)\)
sin o = sin \(\left(-\frac{\pi}{4}\right)\)
Hence θ = \(-\frac{\pi}{4}\) is the principal solution.
The general solution is
θ = nπ + (- 1)ⁿ . \(\left(-\frac{\pi}{4}\right)\) , n ∈ Z
θ = nπ + (- 1)^{n + 1} . \(\frac{\pi}{4}\) , n ∈ Z

(ii) cot θ = √3

The principal value of tan θ lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Since tan θ = \(\frac{1}{\sqrt{3}}\) > 0
The principal value of tan θ lies in the I quadrant.

The general solution of tan θ is
θ = nπ + \(\frac{\pi}{6}\) , n ∈ Z

(iii) tan θ = –\(\frac{1}{\sqrt{3}}\)
The principal value of tan θ lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Since tan θ = – \(\frac{1}{\sqrt{3}}\) > 0
The principal value of tan θ lies in the IV quadrant.

The general solution of tan θ is
θ = nπ – \(\frac{\pi}{6}\) , n ∈ Z

Question 2.
Solve the following equations for which solutions lies in the interval 0° ≤ 9 < 360°
(i) sin⁴x = sin²x
Answer:
sin⁴x – sin²x = 0
sin² x (sin² x – 1) = 0
sin² x [ – (1 – sin² x)] = 0
sin²x × – cos²x = 0
– sin²x cos²x = 0
(sin x cos x)² = 0
(\(\frac { 1 }{ 2 }\) × 2 sin cos x)² = 0
\(\frac { 1 }{ 4 }\) sin 2x = 0
sin 2x = 0
The general solution is
2x = nπ, n ∈ Z
x = \(\frac{\mathrm{n} \pi}{2}\), n ∈ Z

(ii) 2 cos²x + 1 = – 3 cos x
Answer:
2 cos²x + 1 = – 3 cos x
2 cos²x + 3 cos x + 1 = 0
2 cos²x + 2 cos x + cos x + 1 = 0
2 cos x (cos x + 1) + 1 (cos x + 1) = 0
(2 cos x + 1) (cos x + 1) = 0
2 cos x + 1 = 0 or cos x + 1 = 0
cos x = \(-\frac{1}{2}\) or cos x = – 1
To find the solution of cos x = \(-\frac{1}{2}\)
cos x = \(-\frac{1}{2}\)


To find the solution of cos x = – 1
cos x = – 1
cos x = cos π
The general solution is
x = 2nπ ± π, n ∈ Z
x = 2nπ + π or x = 2nπ – π, n ∈ Z

Consider x = 2nπ + π
when n = 0 , x = 0 + π = π ∈ (0°, 360°)
when n = 1 , x = 2π + π = 3π ∉ (0°, 360°)

Consider x = 2nπ – π
when n = 0, x = 0 – π ∉ (0°, 360°)
when n = 1, x = 2π – π = π ∈ (0°, 360°)
when n = 2, x = 4π – π = 3π ∉ (0°, 360°)
∴ The required solution are x = \(\frac{2 \pi}{3}\), \(\frac{4 \pi}{3}\), π

(iii) 2 sin²x + 1 = 3 sin x
Answer:
2 sin²x – 3 sin x + 1 = 0
2 sin²x – 2 sin x – sin x + 1 = 0
2 sin x (sin x – 1) – 1 (sin x – 1) = 0
(2 sin x – 1)(sin x – 1) = 0
2 sin x – 1 = 0 or sin x – 1 = 0
sin x = \(\frac { 1 }{ 2 }\) or sin x = 1
To find the solution of sin x = \(\frac { 1 }{ 2 }\)
sin x = \(\frac { 1 }{ 2 }\)
sin x = sin \(\left(\frac{\pi}{6}\right)\)
The general solution is x = nπ + (-1)ⁿ\(\frac{\pi}{6}\), n ∈ z

(iv) cos²x = 1 – 3 sin x
Answer:
1 – 2 sin²x = 1 – 3 sinx
2 sin² x – 3 sin x = 0
sin x(2 sin x – 3) = 0smx =
sin x = 0 or 2 sin x – 3 = 0
sin x = 0 or sin x = \(\frac{3}{2}\)
sin x = \(\frac{3}{2}\) is not possible since sin x ≤ 1
∴ sin x = 0 = sin 0
The general solution is x = nit ,
When n = 0, x = 0 ∉ (0°, 360°)
When n = 1, x = π ∈ (0°, 360°)
When n = 2, x = 2π ∉ (0°, 360°)
∴ The required solutions is x = π

Question 3.
Solve the following equations:
(i) sin 5x – sin x = cos 3x
Answer:

2 cos 3 x . sin 2x = cos 3 x
2 cos 3x . sin 2x – cos3x = 0
cos 3x (2 sin 2x – 1) = 0
cos 3x = 0 or 2 sin 2x – 1 = 0
cos 3x = 0 or sin 2x = \(\frac { 1 }{ 2 }\)
To find the general solution of cos 3x = 0
The general solution of cos 3x = 0 is
3x = (2n + 1)\(\frac{\pi}{2}\), n ∈ Z
x = (2n + 1)\(\frac{\pi}{6}\), n ∈ Z
To find the general solution of sin 2x = \(\frac{1}{2}\)
sin 2x = \(\frac{1}{2}\)
sin 2x = sin \(\left(\frac{\pi}{6}\right)\)
The general solution is

∴ The required solutions are

(ii) 2 cos²θ + 3 sin θ – 3 = θ
Answer:
2 cos²θ + 3 sin θ – 3 = θ
2(1 – sin²θ)+ 3 sin θ – 3 = θ
2 – 2 sin²θ + 3 sin θ – 3 = θ
– 2 sin²θ + 3 sin θ – 1 = θ
2 sin² θ – 3 sin θ + 1 = θ
2 sin²θ – 2 sin θ – sin θ + 1 = θ
2 sin θ (sin θ – 1) – (sin θ – 1) = θ
(2 sin θ – 1) (sin θ – 1) = 0
2 sin θ – 1 = 0 or sin θ – 1 = θ
sin θ = \(\frac { 1 }{ 2 }\) or sin θ = 1
To find the general solution of’ sin θ = \(\frac { 1 }{ 2 }\)
sin θ = \(\frac { 1 }{ 2 }\)
sin θ = sin \(\frac{\pi}{6}\)
The general solution is θ = nπ + (-1)ⁿ\(\frac{\pi}{6}\), n ∈ Z
To find the general solution of sin θ = 1
sin θ = 1
sin θ = \(\frac{\pi}{2}\)
The general solution is θ = nπ + (-1)ⁿ\(\frac{\pi}{2}\), n ∈ Z
∴ The required solutions are
θ = nπ + (-1)ⁿ\(\frac{\pi}{6}\), n ∈ Z (or)
θ = nπ + (-1)ⁿ\(\frac{\pi}{6}\), n ∈ Z

(iii) cos θ + cos 3θ = 2 cos 2θ
Answer:
cos 3θ + cos θ = 2 cos 2θ

2 cos 2θ . cos θ = 2 cos 2θ
cos 2θ . cos θ – cos 2θ = θ
cos 2θ (cos θ – 1) = θ
cos 2θ = θ or cos θ – 1 = θ
cos 2θ = θ or cos θ = 1
To find the general solution of cos 2θ = θ
The general solution is
2θ = (2n + 1)\(\frac{\pi}{2}\), n ∈ Z
θ = (2n + 1)\(\frac{\pi}{4}\), n ∈ Z
To find the general solution of cos θ = 1
cos θ = 1
cos θ = cos 0
The general solution is θ = 2nπ , n ∈ Z
∴ The required solutions are
θ = (2n + 1)\(\frac{\pi}{4}\), n ∈ Z (or)
θ = 2nπ, n ∈ Z

(iv) sin θ + sin 3θ + sin 5θ = 0
Answer:
sin 5θ + sin 3θ + sin θ = 0

2 sin 3θ . cos 2θ + sin 3θ = 0
sin 3θ (2 cos 2θ + 1) = θ
sin 3θ = 0 or 2 cos 2θ + 1 = θ
sin 3θ = 0 or cos 2θ = –\(\frac { 1 }{ 2 }\)
To find the general solution of sin 3θ = 0
The general solution is
3θ = nπ, n ∈ Z
θ = \(\frac{\mathbf{n} \pi}{3}\), n ∈ Z
To find the general solution of cos 2θ = –\(\frac { 1 }{ 2 }\)

The general solution is

∴ The required solutions are

(v) sin 2θ – cos 2θ – sin θ + cos θ = θ
Answer:

(vi) sin θ + cos θ = √2
Answer:

The general solution is

(vii) sin θ + √3 cos θ = 1
Answer:
Divide each term by 2

(viii) cot θ + cosec θ = √3
Answer:

(ix) tan θ + tan \(\left(\theta+\frac{\pi}{3}\right)\) + tan \(\left(\theta+\frac{2 \pi}{3}\right)\) = √3
Answer:

(x) cos 2θ = \(\frac{\sqrt{5}+1}{4}\)
Answer:
we know cos 36° = \(\frac{\sqrt{5}+1}{4}\), 36° = \(\frac{\pi}{5}\)
cos 2θ = cos 36° = cos \(\left(\frac{\pi}{5}\right)\)
The general solution is
2θ = 2nπ ± \(\frac{\pi}{5}\), n ∈ Z
θ = nπ ± \(\frac{\pi}{10}\), n ∈ Z

(xi) 2cos ²x – 7 cos x + 3 = 0
Answer:
2 cos²x – 7 cos x + 3 = 0
2 cos²x – 6 cos x – cos x + 3 = 0
2 cos x (cos x – 3) – 1 (cos x – 3) = 0
(2 cos x – 1) (cos x – 3) = 0
2 cos x – 1 = 0 or cos x – 3 = 0
cos x = \(\frac { 1 }{ 2 }\) or cos x = 3
Since – 1 ≤ cos x ≤ 1 , we have
cos x = 3 is not possible.
∴ cos x = \(\frac { 1 }{ 2 }\)
cos x = cos \(\frac{\pi}{3}\)
The general solution is x = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.7 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.7

Question 1.
If A + B + C = 180° prove that
(i) sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
Answer:
sin 2A + sin 2 B + sin 2 C = 2 sin \(\left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right)\) . cos \(\left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\) + 2 sin C cos C
= 2 sin (A + B) . cos (A – B) + 2 sin C cos C
= 2 sin( 180° – C) cos (A – B) + 2 sin C cos C (∴ A + B + C = 180°)
= 2 sin C cos (A – B) + 2 sin C cos C
= 2 sin C (cos(A – B) + cos C)
= 2 sin C [(cos (A – B) + cos (180° – (A + B))]
= 2 sin C [cos (A – B) – cos (A + B)]

= 2 sin C . 2 sin A . sin B
= 4 sin A sin B sin C

(ii) cos A + cos B – cos C = -1 + 4 cos \(\frac{\mathbf{A}}{2}\) cos \(\frac{B}{2}\) sin \(\frac{\mathbf{c}}{2}\)
Answer:

(iii) sin 2A + sin 2B + sin 2C = 2 + 2 cos A cos B cos C
Answer:

(iv) sin 2A + sin 2B – sin 2C = 2 sin A sin B cos C
Answer:
Given A + B + C = 180° ⇒ c = 180° – (A + B)

(v)
Answer:
Given A + B + C = 180° ⇒ A + B = 180° – C

(vi) sin A + sin B + sin C = 4 cos \(\frac{\mathbf{A}}{2}\) cos \(\frac{\mathbf{B}}{2}\) cos \(\frac{\mathbf{c}}{2}\)
Answer:

(vii) sin(B + C – A) + sin(C + A – B) + sin(A + B – C) = 4 sin A sin B sin C
Answer:
Given A + B + C = 180°
sin ( B + C – A) = sin (180° – A – A) = sin (180° – 2A) = sin 2A
sin ( C + A – B) = sin (180° – B – B) = sin ( 180° – 2B) = sin 2B
sin (A + B – C) = sin (180° – C – C) = sin ( 180° – 2C) = sin 2C
∴ sin ( B + C – A) + sin ( C + A – B) + sin (A + B – C) = sin 2A + sin 2B + sin 2C
= 2 sin \(\left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right)\) . cos \(\left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\) + 2 sin C . cos C
= 2 sin(A + B) . cos(A – B) + 2 sin C cos C
= 2 sin(180° – C) cos(A – B) + 2 sin C cos C
= 2 sin C cos(A – B) + 2 sin C cos C
= 2sin C(cos(A – B) + cos C)
= 2 sin C [cos(A – B) + cos (180° -(A + B))]
= 2 sin C [cos(A – B) – cos(A + B)]

Question 2.
If A + B + C = 2s, then prove that sin (s – A) sin (s- B )+ sin s . sin(s – C) = sin A sin B
Answer:
Given A + B + C = 2s , we have sin A sin B = \(\frac { 1 }{ 2 }\) [cos ( A – B) – cos ( A + B)]
sin(s – A) sin(s – B) + sins . sin(s – C) = \(\frac { 1 }{ 2 }\) [cos((s – A) – (s – B)) – cos(s – A + s – B)] + \(\frac { 1 }{ 2 }\) [cos (s – (s – C)) – cos (s + s – C)]
= \(\frac { 1 }{ 2 }\) [cos (s – A – s + B) – cos(2s – A – B)] + \(\frac { 1 }{ 2 }\) [cos(s – s + C) – cos(2s – C)]
= \(\frac { 1 }{ 2 }\) [cos (B – A) – cos(2s – A – B)] + \(\frac { 1 }{ 2 }\) [cos C – cos(2s – C)]
= \(\frac { 1 }{ 2 }\) [cos(A – B) – cos(A + B + C – A – B)] + \(\frac { 1 }{ 2 }\) [cos C – cos (A + B + C – C)]
= \(\frac { 1 }{ 2 }\) [cos (A – B) – cos C] + \(\frac { 1 }{ 2 }\) [cos C – cos(A + B)]
= \(\frac { 1 }{ 2 }\) [cos (A – B) – cos C + cos C – cos (A + B)] = \(\frac { 1 }{ 2 }\) [cos(A – B) – cos(A + B)]

Question 3.
If x + y + z = xyz prove that

Answer:
Given x + y + z = x y z ,
Let x = tan A,
y = tan B,
z = tan C
x + y + z = xyz ⇒ tan A + tan B + tan C = tan A tan B tan C
tan A + tan B + tan C – tan A tan B tan C = 0 ——– (1)

tan A + tan B + tan C – tan A tan B tan C = 0 ——– (2)
A + B + C = 180° ⇒ tan A + tan B + tan C – tan A tan B tan C = 0
⇒ tan (A + B + C) = 0
⇒ tan A + tan B + tan C – tan A tan B tan C = 0
⇒ tan (A + B + C) = 0
∴ tan 2(A + B + C) = o
⇒ tan (2A + 2B + 2C) = 0
⇒ tan 2A + tan 2B + tan 2C – tan 2A tan 2B tan 2C = 0 By eqn (1)
tan 2 A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

Question 4.
If A + B + C = prove the following
(i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
Answer:

(ii) cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C
Answer:

Question 5.
If ∆ABC is a right triangle and if ∠A = \(\frac{\pi}{2}\) then prove that
(i) cos² B + cos² C = 1
Answer:

∆ ABC is a right triangle. Given ∠A = 90°
we know A + B + C = 180°
∴ B + C = 180° – A
B + C = 180° – 90° = 90°

(ii) sin² B + sin² C = 1
Answer:

(iii) cos B – cos C = -1 + 2√2 cos \(\frac{\mathbf{B}}{2}\) . sin \(\frac{\mathbf{C}}{2}\)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.6 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.6

Question 1.
Express each of the following as a sum or difference.
(i) sin 35°. cos 28°
(ii) sin 4x cos 2x
(iii) 2 sin 10θ . cos 2θ
(iv) cos 5θ . cos 2θ
(v) sin 5θ . sin 4θ
Answer:
(i) sin 35°. cos 28°
We know
sin A cos B = \(\frac { 1 }{ 2 }\) [sin (A + B) + sin (A – B)]
Take A = 35° and B = 28°
sin 35°cos 28° = \(\frac { 1 }{ 2 }\)[sin(35° + 28°) + sin(35° – 28°)]
sin 350 cos 28° = \(\frac { 1 }{ 2 }\)[sin 63° + sin 7°]

(ii) sin 4x cos 2x
We know
sin A cos B = \(\frac { 1 }{ 2 }\) [sin (A + B) + sin (A – B )]
Take A = 4x , B = 2x
sin 4x . cos 2x = \(\frac { 1 }{ 2 }\)[sin(4x + 2x) + sin(4x – 2x)]
sin 4x . cos 2x = \(\frac { 1 }{ 2 }\)[sin 6x + sin 2x]

(iii) 2 sin 10θ . cos 2θ
We know
2 sin A cos B = sin (A + B) + sin (A – B)
Take A = 10θ, B = 2θ
2 sin 10θ . cos 2θ = sin (10θ + 2θ) + sin (10θ – 2θ)
2 sin 10θ. cos 2θ = sin 12 θ + sin 8θ
2 sin 10θ . cos 2θ = \(\frac { 1 }{ 2 }\)[sin 12θ + sin 8θ]

(iv) cos 5θ . cos 2θ
We know .
cosA cosB = \(\frac { 1 }{ 2 }\) [cos (A + B) + cos (A – B)]
Take A = 5θ, B = 2θ
cos 5θ . cos 2θ = \(\frac { 1 }{ 2 }\) [cos (5θ + 2θ) + cos(5θ – 2θ)]
cos 5θ . cos 2θ = \(\frac { 1 }{ 2 }\) [cos 7θ + cos 3θ]

(v) sin 5θ . sin 4θ
we know
sin A sin B = \(\frac { 1 }{ 2 }\) [cos (A – B) – cos (A + B)]
Take A = 5θ, B = 4θ
sin 5θ . sin 4θ = \(\frac { 1 }{ 2 }\) [cos (5θ – 4θ) – cos (5θ + 4θ)]
sin 5θ . sin 4θ = \(\frac { 1 }{ 2 }\) [cos θ – cos 9θ]

Question 2.
Express each of the following as a product.
(i) sin 75° sin 35°
(ii) cos 65° + cos 15°
(iii) sin 50° + sin 40°
(iv) cos 35° – cos 75°
Answer:
(i) sin 75° sin 35°

(ii) cos 65° + cos 15°

(iii) sin 50° + sin 40°
We know

(iv) cos 35° – cos 75°
We know

Question 3.
Show that sin 12° . sin 48° . sin 54° = \(\frac{1}{8}\)
Answer:
sin 12° . sin 48° . sin 54° = sin 48° . sin 12°. sin (90° – 36°)
= \(\frac { 1 }{ 2 }\) [cos (48° – 12°) – cos (48° + 12°)] cos 36°
= \(\frac { 1 }{ 2 }\) [cos 36° – cos 6o°] cos 36°
= \(\frac { 1 }{ 2 }\) [cos 36° – \(\frac { 1 }{ 2 }\)] cos 36°
= \(\frac { 1 }{ 2 }\) [cos²36° – \(\frac { 1 }{ 2 }\) cos 36°]

Question 4.
Show that

Answer:

Question 5.
Show that

Answer:

Question 6.
Show that

Answer:

Question 7.
Prove that sin x + sin 2x + sin 3x = sin 2x (1 + 2 cos x)
Answer:
sin x + sin 2x + sin 3x = sin x + 2 sin x cos x + 3 sin x – 4 sin³ x
= sin x [1 + 2 cos x + 3 – 4 sin² x]
= sin x [2 cos x + 4 – 4 sin² x ]
= sin x [2 cosx + 4(1 – sin²x)]
= sin x [2 cos x + 4 cos²x]
= 2 sin x cos x [1 + 2 cos x]
= sin 2x (1 + 2 cosx)

Question 8.
Prove that

Answer:

Question 9.
Prove that 1 + cos 2x + cos 4x + cos 6x = 4 cos x . cos 2x . cos 3x
Answer:
4 cos x cos 2x . cos 3x = 4 cos x . cos 3x . cos 2x = 4 cos x . [cos (3x + 2x) + cos (3x – 2x)]
2 cos x. [cos 5x + cos x] = 2 cos 5x . cos x + 2 cos² x
= 2 × \(\frac { 1 }{ 2 }\) [cos (5x + x) + cos (5x – x)] + 1 + cos 2x
= cos 6x + cos 4x + 1 + cos 2x
= 1 + cos 2x + cos 4x + cos 6x

Question 10.
Prove that

Answer:

Question 11.
Prove that cos (30°- A) cos (30° + A) + cos (45° – A). cos(45° + A) = cos 2A + \(\frac { 1 }{ 4 }\)
Answer:
cos(30° – A) cos(30° + A) + cos(45° – A) . cos(45° + A)
= cos (30° + A) cos (30°- A) + cos (45° + A) cos (45° – A)
= \(\frac { 1 }{ 2 }\) [cos (30° + A + 30° – A) + cos ( 30° + A – (30° + A ))] + \(\frac { 1 }{ 2 }\) [cos (45° + A + 45° – A) + cos (45° + A – (450 + A))
= \(\frac { 1 }{ 2 }\) [cos 60° + cos (30° + A – 30° + A)] + \(\frac { 1 }{ 2 }\)[cos 90° + cos(45° + A – 45° + A)]
= \(\frac { 1 }{ 2 }\)[cos 60° + cos 2A] + \(\frac { 1 }{ 2 }\)[cos 90° + 2A]
= \(\frac { 1 }{ 2 }\) cos 60° + \(\frac { 1 }{ 2 }\) cos 2A + \(\frac { 1 }{ 2 }\) cos 90° + \(\frac { 1 }{ 2 }\) cos 2A
= \(\frac { 1 }{ 2 }\) × \(\frac { 1 }{ 2 }\) + cos 2A + \(\frac { 1 }{ 2 }\) × o
= \(\frac { 1 }{ 4 }\) + cos 2A

Question 12.
Show that

Answer:

Question 13.
Prove that

Answer:

Question 14.
Show that cot (A + 15°) – tan (A – 15°) = \(\frac{4 \cos 2 A}{1+2 \sin 2 A}\)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5

Question 1.
Find the value of cos 2A , A lies in the first quadrant when
(i) cos A = \(\frac{15}{17}\)
Answer:
we know sin² A + cos² A = 1
sin² A = 1 – cos² A

Since A lies in the first quadrant, sin A is positive
∴ sin A = \(\frac{8}{17}\)
cos 2A = cos² A – sin² A

(ii) sin A = \(\frac{4}{5}\)
Answer:
we know sin² A + cos² A = 1
cos² A = 1 – sin²A


Since A lies in the first quadrant, cos A is positive
∴ cos A = \(\frac{3}{5}\)
cos 2A = cos² A – sin² A

(iii) tan A = \(\frac{16}{63}\)
Answer:

Question 2.
If θ be an acute angle, find
(i) sin \(\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\), when sin θ = \(\frac{1}{25}\)
(ii) cos \(\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\), when sin θ = \(\frac{8}{9}\)
Answer:
(i) sin \(\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\), when sin θ = \(\frac{1}{25}\)

(ii) cos \(\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\), when sin θ = \(\frac{8}{9}\)

Question 3.
If cos θ =\(\frac{1}{2}\left(a+\frac{1}{a}\right)\), show that cos 3θ = \(\frac{1}{2}\left(a^{3}+\frac{1}{a^{3}}\right)\)
Answer:

Question 4.
Prove that
cos 5θ = 16 cos⁵θ – 20 cos³θ + 5 cos θ
Answer:

Question 5.
Prove that sin 4α = 4 tan α \(\frac{1-\tan ^{2} \alpha}{\left(1+\tan ^{2} \alpha\right)^{2}}\)
Answer:

Question 6.
If A + B = 45°, show that (1 + tan A) (1 + tan B) = 2
Answer:
Given A + B = 45°
tan(A + B) = tan 45°

tan A + tan B = 1 – tan A . tan B —— (1)
(1 + tan A)(1 + tan B) = 1 + tan B + tan A + tan A tan B
= 1 + (tan A + tan B) + tan A tan B
= 1 + 1 – tan A tan B + tan A tan B (By equation (1))
= 2

Question 7.
Prove that (1 + tan 1°) (1 + tan 2°) (1 + tan 3°) …….. (1 + tan 44°) is a multiple of 4.
Answer:
1 + tan 44° = 1 + tan (45° – 1°)

(1 + tan 1°)(1 + tan 44°) = 2
Similarly (1 + tan 2°) (1 + tan 43°) = 2
(1 + tan 3°) (1 + tan 42°) = 2
(1 + tan 22°) (1 + tan 23°) = 2
= (1 + tan 1°) (1 + tan 2°)… (1 + tan 44°) = 2 × 2 × … 22 times
It is a multiple of 4.

Question 8.
Prove that tan \(\left(\frac{\pi}{4}+\theta\right)\) – tan \(\left(\frac{\pi}{4}-\theta\right)\) = 2 tan 2θ
Answer:

Question 9.
Show that

Answer:

Question 10.
prove that (1 + sec 2θ) (1 + sec 4θ) …………….. (1 + sec 2ⁿθ) = tan 2ⁿθ . cot θ.
Answer:

Question 11.
Prove that

Answer:

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:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.3 Text Book Back Questions and Answers, Notes.

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 \(\left(\frac{-11 \pi}{3}\right)\)
Answer:
(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 (1050°) = tan(12 × 90 – 30°)
= – tan30° = – \(\frac{1}{\sqrt{3}}\)

(v) cot 660°
cot 660° = cot (7 × 90 + 30°)
= – tan 30° = – \(\frac{1}{\sqrt{3}}\)

(vi) tan \(\left(\frac{19 \pi}{3}\right)\)

(vii) sin \(\left(\frac{-11 \pi}{3}\right)\)

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

Question 3.
Find the values of the other five trigonometric functions of the following
(i) cos θ = –\(\frac{1}{2}\), θ lies in the III quadrant
(ii) cos θ = \(\frac{2}{3}\), θ lies in the I quadrant
(iii) sin θ = –\(\frac{2}{3}\), θ lies in the IV quadrant
(iv) tan θ = – 2, θ lies in the II quadrant
(v) sec θ = \(\frac{13}{5}\), θ lies inthe IVquadrant
Answer:
(i) cos θ = –\(\frac{1}{2}\), θ lies in the III quadrant

(ii) cos θ = \(\frac{2}{3}\), θ lies in the I quadrant
We know that cos²θ + sin²θ = 1
\(\left(\frac{2}{3}\right)^{2}\) + sin²θ = 1
\(\frac{4}{9}\) + sin²θ = 1

Since θ lies in the I quadrant all trigonometric functions are positive.

(iii) sin θ = –\(\frac{2}{3}\), θ lies in the IV quadrant
We know that cos²θ + sin²θ = 1
cos²θ + \(\left(-\frac{2}{3}\right)^{2}\) = 1
cos²θ + \(\frac{4}{9}\) = 1

Since θ lies in the fourth quadrant cos θ is positive.

(iv) tan θ = – 2, θ lies in the II quadrant
We know that sec²θ – tan²θ = 1
sec²θ – (-2)² = 1
sec²θ – 4 = 1
sec²θ = 1 + 4 = 5
sec θ = ± √5
Since θ lies in the second quadrant sec θ is negative.

(v) sec θ = \(\frac{13}{5}\), θ lies inthe IVquadrant
We know that sec²θ – tan²θ = 1
\(\left(\frac{13}{5}\right)^{2}\) – tan²θ = 1
\(\frac{169}{25}\) – 1 = tan²θ

Question 4.
Prove that

Answer:

Question 5.
Find all the angles between 0° and 360° which satisfy the equation sin²θ = \(\frac{3}{4}\)
Answer:
sin²θ = \(\frac{3}{4}\) ⇒ sin θ = ± \(\frac{\sqrt{3}}{2}\)
sin 60° = \(\frac{\sqrt{3}}{2}\)
sin 120° = sin (180° – 60°)
= sin 60° = \(\frac{\sqrt{3}}{2}\)
∴ θ = 60° and 120°

Question 6.
Show that

Answer:

= sin² 10° + sin² 20° + [cos 20°]² + [cos 10°]²
= sin² 10° + sin² 20° + cos² 20° + cos² 10°
= sin² 10° + cos² 10° + sin² 20° + cos² 20°
= 1 + 1 = 2

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.2

Question 1.
Express each of the following in radian measure.
(i) 30°
(ii) 135°
(iii) -205°
(iv) 150°
(v) 330°
Answer:
(i) 30°

(ii) 135°

(iii) – 205°

(iv) 150°

(v) 330°

Question 2.
Find the degree measure corresponding to the following radian measures.
(i) \(\frac{\pi}{3}\)
(ii) \(\frac{\pi}{9}\)
(iii) \(\frac{2 \pi}{5}\)
(iv) \(\frac{7 \pi}{3}\)
(v) \(\frac{10 \pi}{9}\)
Answer:
(i) \(\frac{\pi}{3}\) radians

(ii) \(\frac{\pi}{9}\) radians

(iii) \(\frac{2 \pi}{5}\) radians

(iv) \(\frac{7 \pi}{3}\) radians

(v) \(\frac{10 \pi}{9}\) radians

Question 3.
What must be the radius of a circular running path, around which an athlete must run 5 times in order to describe 1 km?
Answer:
Let the radius of the circular path be = r m.
Length of the circular path s = 1 k. m
s = 1000 m.
Athlete runs 5 times around the path to cover 1 k. m distance
∴ θ = 360° × 5
θ = 360° × 5 × \(\frac{\pi}{180}\) radians
θ = 10 π radians
s = r θ
1000 = r 10 π
r = \(\frac{1000}{10 \pi}\)
r = \(\frac{1000 \times 7}{10 \times 22}=\frac{350}{11}\)
r = 31 .818 meters
Radius of the circular path = 31 .82 meters

Question 4.
In a circle of diameter 40 cm a chord is of length 20 cm. Find the length of the minor arc of the chord.
Answer:

Given Diameter AB = 40 cm
∴ Radius r = 20 cm
Chord CD = 20 cm
O – Centre of the circle
OC = OD = radius = 20 cm.
∴ Triangle OCD is an equilateral triangle.

To find the length of the minor arc CD.
Let s = minor arc CD.
The arc CD subtends 60° at the centre.
θ = 60°
θ = 60° × \(\frac{\pi}{180}\) radians.
θ = \(\frac{\pi}{3}\) radians
We have s = rθ

Question 5.
Find the degree measure of the angle subtended at the centre of the circle of radius 100 cm by an arc of length 22 cm.
Answer:
Given radius r = 100 cm.
Length of arc s = 22 cm.
Angle subtended by the arc at the centre = θ radians

Question 6.
What is the length of the arc intercepted by a central angle of measure 41° in a circle of radius of 10 feet?
Answer:
Central angle subtended by the arc θ = 41°
θ = 41 × \(\frac{\pi}{180}\) Radians
The radius of the circle r = 10 feet
Length of the arc = s
s = rθ

Question 7.
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Answer:
Let r₁ and r₂ be the radii of the two circles and l be the length of the arc.
Given central angle θ₁ = 60°

Question 8.
The perimeter of a certain sector of a circle is equal to the length of the arc of a semi-circle having the same radius. Express the angle of the sector in degrees, minutes, and seconds.
Answer:
Let OAB be the sector of a circle of radius r.
The angle of the sector is θ.

Perimeter of the sector = OA + arc AB + OB arc AB = rθ
∴ Perimeter of the sector = r + r θ + r
= 2r + rθ
= r(2 + θ) ———- (1)
Length of the arc of the semi – circle of radius
l = nπ ——– (2)
Given that perimeter the circular sector = Length of the arc of the semi circle of radius r
From equations (1) and (2), we have
r(2 + θ) = πr
2 + θ = π
θ = π – 2

Question 9.
An airplane propeller rotates 1000 times per minute. Find the number of degrees that a point on the edge of the propeller will rotate in 1 second.
Answer:
Given An airplane, the propeller rotates 1000 times per minute.
∴ A point on the edge of the propeller also rotates 1000 times in 1 minute.
∴ In 1 minute the point describes 1000 × 2π radians angle at the centre.
In 60 seconds the point describes 1000 × 2π radians angle.
∴ In 1 second the angle described = \(\frac{1000 \times 2 \pi}{60}\) radians

Question 10.
A train is moving on a circular track of a 1500 m radius at the rate of 66 km/hr. What angle will it turn in 20 seconds?
Answer:
Radius of the circular track r = 1500 m.
Speed of the train = 66 km/hr
Let θ be the angle made by the path of train at the centre in 20 seconds.
In 1 hr distance moved by train along the circular path = 66 km
In 60 × 60 seconds distance moved = 66 km
∴ In 20 seconds distance moved s = \(\frac{66}{60 \times 60}\) × 20

Question 11.
A circular metallic plate of radius 8 cm and thickness 6 nuns is melted and molded into a pie (a sector of the circle with thickness) of radius 16 cm and thickness 4 mm. Find the angle of the sector.
Answer:
Radius of the circular metallic plate r = 8 cm
Thickness of the plate h = 6 mm = \(\frac{6}{10}\)
Radius of the Pie l = 16 cm
Thickness of the Pie ( h) = 4mm = \(\frac{4}{10}\) cm
Given volume of the cylinder = Volume of the sector

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.12 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12

Question 1.
Let b > 0 and b ≠ 1. Express y = b^x in logarithmic form. Also, state the domain and range of the logarithmic function.
Answer:
Given y = b^x ⇒ log_by = x, x ∈ R with range (0, ∞) (-∞, ∞)

Question 2.
Compute log₉ 27 – log₂₇ 9
Answer:
log₉27 – log₂₇9 = log₉ 3³ – log₂₇ 3²
= 3 log₉ 3 – 2 log₂₇ 3 —— (1)
By change of base rule [log_b a = \(\frac{1}{\log _{a} b}\)]

Question 3.
Solve log_ax + log₄x + log₂x = 11
Answer:

Question 4.
Solve log ₄ 2 ^8x = 2 ^log₂8
Answer:
Given log ₄ 2^8x = 2 ^log₂8
log ₄ 2^8x = 2^log₂23
log ₄ 2^8x = 2^{3 log₂2}
log ₄ 2^8x = 2³ = 8
2^8x = 4⁸
(2²)^4x = 4⁸
⇒ (4)^4x = 4⁸
⇒ 4x = 8
⇒ x = \(\frac { 8 }{ 4 }\) = 2

Question 5.
If a² + b² = 7ab, show that
log \(\left(\frac{a+b}{3}\right)\) = \(\frac{1}{2}\) (log a + log b)
Answer:
Given
a² + b² = 7ab
Adding both sides 2ab we get
a² + b² + 2ab = 7ab + 2ab
(a + b)² = 9ab

Taking square root on both sides

Taking logarithm on both sides

Question 6.
Prove that log \(\frac{\mathbf{a}^{2}}{\mathbf{b c}}\) + log \(\frac{\mathbf{b}^{2}}{\mathbf{c a}}\) + log \(\frac{c^{2}}{a b}\) = 0
Answer:

Question 7.
Prove that

Answer:

Question 8.
Prove that log_a² a + log_b² b + log_c² c = \(\frac{1}{8}\)
Answer:

Question 9.
Prove log a + log a² + log a³ + ……… + log aⁿ = \(\frac{n(n+1)}{2}\) log a
Answer:
log a + log a² + log a³ + ……… + log aⁿ
= log a + 2 log a + 3 log a + ………….. + n log a
= log a (1 + 2 + 3 + ………….. + n)
= log a × \(\frac{n(n+1)}{2}\)
= \(\frac{n(n+1)}{2}\) log a

Question 10.
If , then prove that xyz = 1
Answer:
Let \(\frac{\log x}{y-z}\) = k
log x = k(y – z)
log x = ky – kz ——— (1)
Similarly log y = k(z – x) = kz – kx ——(2)
log z = k(x – y) = kx – ky ——- (3)
Adding (1), (2) and (3)
log x + log y + log z = ky – kz + kz – kx + kx – ky
log (xyz) = 0 = log 1
xyz = 1

Question 11.
Solve log₂x – 3 log_1/2x = 6
Answer:

Question 12.
Solve log_{5 – x} (x² – 6x + 65) = 2
Answer:
log_{5 – x}(x² – 6x + 65) = 2
⇒ x² – 6x + 65 = (5 – x)²
⇒ x² – 6x + 65 = 25 + x² – 10x
⇒ x² – 6x + 65 – 25 – x² + 10x = 0
⇒ 4x + 40 = 0
⇒ 4x = -40
⇒ x = -10