Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

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)]

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 2
= 2 sin C . 2 sin A . sin B
= 4 sin A sin B sin C

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

(ii) cos A + cos B – cos C = -1 + 4 cos \(\frac{\mathbf{A}}{2}\) cos \(\frac{B}{2}\) sin \(\frac{\mathbf{c}}{2}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 3

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

(iii) sin 2A + sin 2B + sin 2C = 2 + 2 cos A cos B cos C
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 4

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

(iv) sin 2A + sin 2B – sin 2C = 2 sin A sin B cos C
Answer:
Given A + B + C = 180° ⇒ c = 180° – (A + B)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 5
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 6

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

(v) Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 1
Answer:
Given A + B + C = 180° ⇒ A + B = 180° – C
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 7

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

(vi) sin A + sin B + sin C = 4 cos \(\frac{\mathbf{A}}{2}\) cos \(\frac{\mathbf{B}}{2}\) cos \(\frac{\mathbf{c}}{2}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 8

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

(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)]
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 9

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

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)]
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 10

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

Question 3.
If x + y + z = xyz prove that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 11
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)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 12
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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 13

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

Question 4.
If A + B + C = prove the following
(i) sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 14

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

(ii) cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 15
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 16

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

Question 5.
If ∆ABC is a right triangle and if ∠A = \(\frac{\pi}{2}\) then prove that
(i) cos2 B + cos2 C = 1
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 17
∆ ABC is a right triangle. Given ∠A = 90°
we know A + B + C = 180°
∴ B + C = 180° – A
B + C = 180° – 90° = 90°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 18

(ii) sin2 B + sin2 C = 1
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 19

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7

(iii) cos B – cos C = -1 + 2√2 cos \(\frac{\mathbf{B}}{2}\) . sin \(\frac{\mathbf{C}}{2}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.7 20

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

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°]

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

(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]

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

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°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 1

(ii) cos 65° + cos 15°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 2

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

(iii) sin 50° + sin 40°
We know
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 3

(iv) cos 35° – cos 75°
We know
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 4

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

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°]
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 5
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 6

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

Question 4.
Show that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 7
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 8

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

Question 5.
Show that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 9
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 10

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

Question 6.
Show that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 11
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 12

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)

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

Question 8.
Prove that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 13
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 14

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

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

Question 10.
Prove that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 15
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 16

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

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

Question 12.
Show that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 17
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 18

Question 13.
Prove that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 19
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 20

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6

Question 14.
Show that cot (A + 15°) – tan (A – 15°) = \(\frac{4 \cos 2 A}{1+2 \sin 2 A}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.6 21

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°

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 1
25° First quadrant

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

(ii) 825°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 2
825° = 9 × 90° + 15°
825° = 2 × 360° + 105°
∴ 825° lies in the second quadrant.

iii) -55°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 3
-55° lies in the fourth quadrant

iv) 328°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 4
328° = 270° + 58° lies in the fourth quadrant.

v) -230°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 5
– 230° = – 180° + (- 50°) lies in the second quadrant.

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

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°

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 6

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

Question 5.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 7
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 8
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 9

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 10
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 11
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 12

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

Question 6.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 13
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 14
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 15

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

Question 7.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 16
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 17
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 18
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 19

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 20
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 21
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]

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

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 + \(\frac{1}{p}\)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 22

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

Question 10.
If cot θ(1 + sin θ) = 4m and cot θ (1 – sin θ) = 4n then prove that (m2 – n2)2 = mn.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 23

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 24

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

Question 11.
If cosec θ – sin θ = a3, sec θ – cos θ = b3 then prove that a2b2(a2 + b2) = 1.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 25
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 26

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1

Question 12.
Eliminate θ from the equations a sec θ – c tan θ = b, b sec θ + d tan θ = c.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.1 27

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

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

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 1

(vii) sin \(\left(\frac{-11 \pi}{3}\right)\)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 2

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

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.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 3
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 4
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 4

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 6

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

(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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 7
Since θ lies in the I quadrant all trigonometric functions are positive.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 8
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 9

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

(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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 10
Since θ lies in the fourth quadrant cos θ is positive.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 11

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

(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.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 12

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

(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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 13

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

Question 4.
Prove that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 14
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 15
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 16

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

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°

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3

Question 6.
Show that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 17
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.3 18
= 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

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

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°

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 1

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

(ii) 135°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 2
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 3

(iii) – 205°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 4

(iv) 150°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 5

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

(v) 330°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 6

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 7
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 8

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

(ii) \(\frac{\pi}{9}\) radians
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 9

(iii) \(\frac{2 \pi}{5}\) radians
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 10

(iv) \(\frac{7 \pi}{3}\) radians
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 11

(v) \(\frac{10 \pi}{9}\) radians
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 12

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

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

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

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:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 13

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.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 14
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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 15

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 16

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 17

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

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°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 18

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

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.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 19
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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 20
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 21

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 22

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 23

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.2 24

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

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

Samacheer Kalvi 11th Maths Guide 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)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 1
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 2

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

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°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 3
A + 45° = 120°
A = 120° – 45° = 75°
A = 75°

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

Question 3.
In a ∆ ABC, if cos c = \(\frac{\sin \mathbf{A}}{2 \sin B}\) show that the triangle is isosceles.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 4
a2 + b2 – c2 = a2
b2 – c2 = 0
b2 = c2 ⇒ b = c
Two sides of is ∆ ABC are equal.
∴ ∆ ABC is an isosceles triangle.

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

Question 4.
In a ∆ ABC, prove that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 5
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 6
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 7

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

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]
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.9 55
= 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

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

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°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 10

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

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:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 14
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 15

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

(ii) a (cos B + cos C) = 2(b + c) sin2\(\frac{\mathbf{A}}{2}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 16
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 17
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 18
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 19

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

(iii) Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 11
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 20

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

(iv) Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 12
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 21
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 22
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 23

 

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 24

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

(v) Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 13
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 25
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 26

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

Question 8.
In a triangle ∆ ABC, prove that
(a2 – b2 + c2) tan B = (a2 + b2 – c2) tan C
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 27
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 28

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

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.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 29
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 30

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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 31
∴ The dimensions of the triangle are 14 m, 14 m, 14 m.
Maximum area = 49√3 sq.m.

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

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,
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 32
(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

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9

(ii) Using Law of cosines.
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 33
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 34
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.9 35

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12

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 = bx in logarithmic form. Also, state the domain and range of the logarithmic function.
Answer:
Given y = bx ⇒ logby = x, x ∈ R with range (0, ∞) (-∞, ∞)

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12

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 [logb a = \(\frac{1}{\log _{a} b}\)]
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 1

Question 3.
Solve logax + log4x + log2x = 11
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 2

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12

Question 4.
Solve log 4 2 8x = 2 log28
Answer:
Given log 4 28x = 2 log28
log 4 28x = 2log223
log 4 28x = 23 log22
log 4 28x = 23 = 8
28x = 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
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 3
Taking square root on both sides
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 4
Taking logarithm on both sides
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 5

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12

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:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 6

Question 7.
Prove that
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 7
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 8

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12

Question 8.
Prove that loga2 a + logb2 b + logc2 c = \(\frac{1}{8}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 9

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 Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 10, 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

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12

Question 11.
Solve log2x – 3 log1/2x = 6
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 11
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12 12

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.12

Question 12.
Solve log5 – x (x2 – 6x + 65) = 2
Answer:
log5 – 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

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.13 Text Book Back Questions and Answers, Notes.

   

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13

Question 1.
If |x + 2| ≤ 9 then x belongs to
(1) (- ∞,- 7)
(2) [- 11, 7]
(3) (-∞, – 7) ∪ [11, ∞]
(4) (-11, 7)
Answer:
(2) [- 11, 7]

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Explanation:
-x – 2 ≤ 9 x + 2 ≤ 9
-x < 9 + 2 = 11 x ≤ 9 – 2 = 7
⇒ x ≥ -11
so x ∈ [-11, 7]

Question 2.
Given that x, y and b are real numbers x < y, b > 0 then
(1) xb < yb
(2) xb > yb
(3) xb < yb
(4) \(\frac{x}{b}\) ≥ \(\frac{y}{b}\)
Answer:
(1) xb < yb

Explanation:
Given x, y and b are real numbers and b ≥ 0. x > b
Multiplying by positive real number the inequality is not affected.
∴ xb < yb

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 3.
If \(\frac{|x-2|}{x-2}\) ≥ 0 , then x belongs to
(1) [2, ∞)
(2) (2, ∞)
(3) (-∞, 2)
(4) (-2, ∞)
Answer:
(1) [2, ∞)

Explanation:
Given \(\frac{|x-2|}{x-2}\) ≥ 0
By the definition of mod function
|x – 2 | = – (x – 2) if x – 2 < 0 | x – 2 | = x – 2 if x – 2 > 0
Suppose x – 2 < 0 then
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13 1

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 4.
The solution of 5x – 1 < 24 and 5x + 1 > – 24 is
(1) (4, 5)
(2) (- 5, – 4)
(3) (- 5, 5)
(4) (- 5, 4)
Answer:
(3) (- 5, 5)

Explanation:
The given inequalities are
5x – 1 < 24 ——— (1)
5x + 1 > – 24 ——– (2)
(1) ⇒ 5x – 1 < 24
⇒ 5x < 24 + 1
⇒ 5x < 25 ⇒ x < 5 ——— (3)
(2) ⇒ 5x + 1 > – 24
⇒ 5x > – 24 – 1
⇒ 5x > – 25
⇒ x > – 5 ——– (4)
Combining (3) and (4), we have
-5 < x < 5
∴ x ∈ (- 5, 5)

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 5.
The solution set of the following inequality
|x – 1| ≥ |x – 3| is
(1) [0, 2]
(2) [2, ∞)
(3) (0, 2)
(4) (-∞, 2)
Answer:
(2) [2, ∞)

Explanation:
The given inequality is |x – 1| ≥ |x – 3|
(x – 1)2 ≥ (x – 3)2
x2 – 2x + 1 ≥ x2 – 6x + 9
– 2x + 1 ≥ – 6x + 9
6x – 2x + 1 – 9 ≥ 0
4x – 8 ≥ 0 ⇒ 4x ≥ 8
⇒ x ≥ 2
∴ The solution set of the given inequality lies in the interval (2, ∞)

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 6.
The value of \(\log _{\sqrt{2}} 512\) is
(1) 16
(2) 18
(3) 9
(4) 12
Answer:
(2) 18

Explanation:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13 2

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 7.
The value of \(\log _{3} \frac{1}{81}\) is
(1) – 2
(2) – 8
(3) – 4
(4) – 9
Answer:
(3) – 4

Explanation:
\(\log _{3} \frac{1}{81}\) = log31 – log381
= o – log3 34
= – 4 log33
= – 4 × 1
= – 4

Question 8.
If log√x0.25 = 4, then the value of x is
(1) 0.5
(2) 2.5
(3) 1.5
(4) 1.25
Answer:
(1) 0.5

Explanation:
log√x0.25 = 4
By the definition of logarithm
0.25 = \((\sqrt{x})^{4}\)
(0.5 )2 = \(\left(x^{\frac{1}{2}}\right)^{4}\)
(0.5 )2 = x2
x = 0.5

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 9.
The value of logab . logbc . logca is
(1) 2
(2) 1
(3) 3
(4) 4
Answer:
(2) 1

Explanation:
logab . logbc . logca = logac . logca
= logaa = 1

Question 10.
If 3 is the logarithm of 343 then, the base is
(1) 5
(2) 7
(3) 6
(4) 9
Answer:
(2) 7

Explanation:
⇒ logx343 = 3 ⇒ 343 = x3
(.i.e.,) 73 = x3 ⇒ x = 7
⇒ x = 7

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 11.
Find a so that the sum and product of the roots of the equation 2x2 + (a – 3 ) x + 3a – 5 = 0 are equal is
(1) 1
(2) 2
(3) 0
(4) 4
Answer:
(2) 2

Explanation:
Given quadratic equation is
2x2 + (a – 3) x + 3a – 5 = 0
Let the roots be α, β
Sum of the roots α + β = \(-\frac{(a-3)}{2}\)
Product of the roots α β = \(\frac{3 a-5}{2}\)
Given α + β = α β
∴ \(-\frac{(a-3)}{2}=\frac{3 a-5}{2}\)
– a + 3 = 3a – 5
3a + a = 5 + 3
4a = 8 ⇒ a = 2

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 12.
If a and b are the roots of the equation x2 – kx +16 = 0 and satisfy a2 + b2 = 32, then the value of k is
(1) 10
(2) – 8
(3) – 8, 8
(4) 6
Answer:
(3) – 8, 8

Explanation:
a + b = k ….(1) ab = 16 ….(2)
a2 + b2 = (a + b)2 – 2ab = 32 .
k2 – 32 = 32 ⇒ k2 = 64 ⇒ k = ±8

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 13.
The number of solutions of x2 + |x – 1| = 1 is
(1) 1
(2) 0
(3) 2
(4) 3
Answer:
(3) 2

Explanation:
The given quadratic equatiuon is
x2 + |x – 1| = 1 ——– (1)

Case(i)
By the definition of mod function if x – 1 ≥ 0, then
|x – 1| = x – 1
∴ (1) ⇒ x2 + x – 1 = 1
x2 + x – 2 = 0
x2 + 2x – x – 2 = 0
x(x + 2) – 1 (x + 2) = 0
(x – 1)(x + 2) = 0
x – 1 = 0 or x + 2 = 0
x = 1 or x = – 2
Since x – 1 ≥ 0
x = – 2 is not possible. ∴ x = 1

Case (ii)
Again by the definition of mod function if x – 1 < 0 then
|x – 1| = – (x – 1)
∴ (1) ⇒ x2 – (x – 1) = 1
x2 – x + 1 = 1
x2 – x = 0
x (x – 1 ) = 0
x = 0 or x – 1 = 0
x = 0 or x = 1
Since x < 1, x = 1 is not possible
∴ x = 0
∴ The required solution set is {0, 1}
Number of solutions = 2

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 14.
The equation whose roots are numerically equal but opposite in sign to the roots of 3x2 – 5x – 7 = 0 is
(1) 3x2 – 5x – 7 = 0
(2) 3x2 + 5x – 7 = 0
(3) 3x2 – 5x + 7 = 0
(4) 3x2 + x – 7 = 0
Answer:
(2) 3x2 + 5x – 7 = 0

Explanation:
The given quadratic equation is
3x2 – 5x – 7 = 0 ——— (1)
Let α and β be the roots of eqn (1)
Sum of the roots α + β = \(-\left(\frac{-5}{3}\right)\)
α + β = \(\frac{5}{3}\)
Product of the roots α β = – \(\frac{7}{3}\)
The quadratic equation whose roots are – α and – β is
x2 – (sum of the roots) x + Product of the roots = 0
x2 – (- α – β)x + (- α)(- β) = 0
x2 + (α + β)x + αβ = 0
x2 + \(\frac{5}{3}\) x – \(\frac{7}{3}\) = 0
3x2 + 5x – 7 = 0 is the required equation.

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 15.
If 8 and 2 are the roots of x2 + ax + c = 0 and 3, 3 are the roots of x2 + dx + b = 0 , then the roots of the equation x2 + ax + b = 0 are
(1) 1, 2
(2) -1, 1
(3) 9, 1
(4) -1, 2
Answer:
(3) 9, 1

Explanation:
Given that 8 and 2 are the roots of the equation
x2 + ax + c = 0 ———- (1)
Sum of the roots 8 + 2 = \(-\frac{a}{1}\) ⇒ a = -10
Product of the roots 8 × 2 = \(\frac{c}{1}\) ⇒ c = 16
Also given 3 , 3 are the roots of
x2 + dx + b = 0 ——— (2)
Sum of the roots 3 + 3 = – \(\frac{d}{1}\) ⇒ d = – 6
Product of the roots 3 × 3 = \(\frac{b}{1}\) ⇒ b = 9
Let a and p be roots of the equation x2 + ax + b = 0
Sum of the roots α + β = – y
α + β = -(-10)
α + β = 10 ——– (3)
Product of the roots α β = y
α β = 9 ———- (4)
(α – β) = (α + β)2 – 4αβ
= 102 – 4 × 9
= 100 – 36 = 64
α – β = 8 ——— (5)
Solving equations (3) and (5)
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13 3
Substituting in equation (3),
9 + β = 10
⇒ β = 10 – 9 = 1
∴ The required roots are 9, 1

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 16.
If a and b are the real roots of the equation x2 – kx + c = 0 , then the distance between the
points(a, 0) and (b, 0)is
(1) \(\sqrt{\mathbf{k}^{2}-4 \mathbf{c}}\)
(2) \(\sqrt{4 k^{2}-c}\)
(3) \(\sqrt{4 \mathbf{c}-\mathbf{k}^{2}}\)
(4) \(\sqrt{\mathbf{k}-8 \mathbf{c}}\)
Answer:
(1) \(\sqrt{\mathbf{k}^{2}-4 \mathbf{c}}\)

Explanation:
Given that a and b are the roots of the equation
x2 – kx + c = 0 ——— (1)
Sum of the roots a + b = \(-\frac{(-k)}{1}\)
a + b = k ———- (2)
Product of the roots ab = \(\frac{c}{1}\)
ab = c ——– (3)
Distance between the points ( a, 0) and (b, 0) is
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13 4

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 17.
If Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13 5 then the value of k is
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3

Explanation:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13 6
kx = 2x – 2 + x + 2
kx = 3x ⇒ k = 3

Question 18.
If Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13 7, then the value of A + B is
(1) – \(\frac{1}{2}\)
(2) – \(\frac{2}{3}\)
(3) \(\frac{1}{2}\)
(4) \(\frac{2}{3}\)
Answer:
(1) – \(\frac{1}{2}\)

Explanation:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13 8
1 – 2x = A (x + 1) + B(3 – x) ——— (1)
Put x = – 1 in equation (1)
1 – 2 (- 1) = A (- 1 + 1) + B (3 + 1)
1 + 2 = 0 + 4B ⇒ B = \(\frac{3}{4}\)
Put x = 3 in equation (1)
1 – 2 × 3 = A(3 + 1) + B(3 – 3)
1 – 6 = 4A + 0
– 5 = 4A
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13 9

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Question 19.
The number of roots of (x + 3)4 + (x + 5)4 = 16 is
(1) 4
(2) 2
(3) 3
(4) 0
Answer:
(1) 4

Explanation:
The equation is (x + 3)4 + (x + 5)4 = 16
(x + 3)4 + (x + 5)4 = 24
This is biquadratic equation. It has 4 roots.

Question 20.
The value of
log3 11 . log11 13 . log 13 15 . log 15 27 . log 27 81 is
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(4) 4

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.13

Explanation:
log3 11 . log11 13 . log 13 15 . log 15 27 . log 27 81
= log3 13 . log 13 15 . log 15 27 . log 27 81
= log 3 15 . log 15 27 . log 27 81
= log 3 27 . log 27 81
= log 3 81
= log 334
= 4 log 33
= 4 × 1
= 4

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

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

Samacheer Kalvi 11th Maths Guide 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
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 1
Since A lies in the first quadrant, sin A is positive
∴ sin A = \(\frac{8}{17}\)
cos 2A = cos2 A – sin2 A
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 2

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

(ii) sin A = \(\frac{4}{5}\)
Answer:
we know sin2 A + cos2 A = 1
cos2 A = 1 – sin2A
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 3
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 4
Since A lies in the first quadrant, cos A is positive
∴ cos A = \(\frac{3}{5}\)
cos 2A = cos2 A – sin2 A
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 5

(iii) tan A = \(\frac{16}{63}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 6

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

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 = \(\frac{1}{25}\)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 7
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 8

(ii) cos \(\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\), when sin θ = \(\frac{8}{9}\)
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 9

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

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:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 10
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 11

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

Question 4.
Prove that
cos 5θ = 16 cos5θ – 20 cos3θ + 5 cos θ
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 12

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

Question 5.
Prove that sin 4α = 4 tan α \(\frac{1-\tan ^{2} \alpha}{\left(1+\tan ^{2} \alpha\right)^{2}}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 13

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

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°
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 14
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

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

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°)
Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5 23
(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.

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

Question 8.
Prove that tan \(\left(\frac{\pi}{4}+\theta\right)\) – tan \(\left(\frac{\pi}{4}-\theta\right)\) = 2 tan 2θ
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 15

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

Question 9.
Show that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 16
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 17

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 18
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 19

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

Question 10.
prove that (1 + sec 2θ) (1 + sec 4θ) …………….. (1 + sec 2nθ) = tan 2nθ . cot θ.
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 20

Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5

Question 11.
Prove that
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 21
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 22
Samacheer Kalvi 11th Maths Guide Chapter 3 Trigonometry Ex 3.5 23

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.11 Text Book Back Questions and Answers, Notes.

   

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.11

Question 1.
Simplify
(a) (125)2/3
(b) 16-3/4
(c) (- 1000)-2/3
(d) (3-6)1/3
(e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)
Answer:
(a) (125)2/3

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 1

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

(b) 16-3/4
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 2

(c) (- 1000)-2/3
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 3

(d) (3-6)1/3
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 4

(e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 5

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Question 2.
Evaluate Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 6
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 7

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Question 3.
If \(\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right)^{2}=\frac{9}{2}\) then find the value of \(\left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right)\) for x > 1
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 8
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 9

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Question 4.
Simplify and hence find the value of n:
\(\frac{3^{2 n} \cdot 9^{2} \cdot 3^{-n}}{3^{3 n}}\) = 27
Answer:
Given
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 10
4 – 2n = 3
2n = 4 – 3
2n = 1
n = \(\frac { 1 }{ 2 }\)

Question 5.
Find the radius of the spherical tank whose volume is \(\frac{32 \pi}{3}\) units.
Answer:
Let r be the radius of the spherical tank.
Given volume of the spherical tank = \(\frac{32 \pi}{3}\)
\(\frac{4}{3}\)πr3 = \(\frac{32 \pi}{3}\)
4r3 = 32
r3 = \(\frac{32}{4}\) = 8
r3 = 23
r = 2
∴ Radius of the spherical tank r = 2 units.

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Question 6.
Simplify by rationalizing the denominator \(\frac{7+\sqrt{6}}{3-\sqrt{2}}\)
Answer:
\(\frac{7+\sqrt{6}}{3-\sqrt{2}}\)
Multiply the numerator and denominator by 3 + √2
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 11

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Question 7.
Simplify:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 12
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 13

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 14
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 15

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Question 8.
If x = √2 + √3 find \(\frac{x^{2}+1}{x^{2}-2}\)
Answer:
Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 16

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11

Samacheer Kalvi 11th Maths Guide Chapter 2 Basic Algebra Ex 2.11 17