Students can Download Basic Maths Exercise 14.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3

Part – A

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 One Marks Questions and Answers

Question 1.
Express each of the following a sum or difference of two trigonometric functions.
(i) sin 5A · cos3A
(ii) cos 4A sin 2A
(iii) cos \(\left(\frac{5 \theta}{2}\right)\) . sin \(\left(\frac{\theta}{2}\right)\)
(iv) 2 cos 70° cos 10°
Answer:
(i) Using transformation formulae
sin5A . cos3A = \(\frac { 1 }{ 2 }\)[sin(5A +3A) + sin(5A – 3A)]
= \(\frac { 1 }{ 2 }\)(sin 8A + sin 2A)

(ii) cos4A – sin2A = \(\frac { 1 }{ 2 }\) [sin 6A – sin 2A]

(iii) cos \(\left(\frac{5 \theta}{2}\right)\) . sin \(\left(\frac{\theta}{2}\right)\) = \(\frac { 1 }{ 2 }\)[sin 3θ – sin 2θ]

(iv) 2cos 70° . cos10° = 2 x \(\frac { 1 }{ 2 }\) [cos80° + cos60°]
= cos80° + \(\frac { 1 }{ 2 }\)

KSEEB Solutions

Question 2.
Express each of the following as the product of two trigonometric reactions.
(i) sin 12x + sin 4x
(ii) sin 7A – sin 3A
(iii) cos 2θ + cos 6θ
(iv) sin 80° – sin 40°
Answer:
(i) sin12x + sin4x = 2.sin \(\frac{12 x+4 x}{2}\) . cos \(\frac{12 x-4 x}{2}\)

(ii) sin7A – sin3A = 2 cos 5A.sin2A

(iii) cos6θ + cos2θ = 2 cos4θ.cos2θ

(iv) sin 80° – sin 40° = 2 cos60°. sin 20° = 2.\(\frac { 1 }{ 2 }\)sin 20° = sin 20°

Part – B

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3Two Marks Questions and Answers

Prove the following

Question 1.
\(\frac{\cos 2 A-\cos 12 A}{\sin 12 A-\sin 2 A}=\tan 7 A\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 1

Question 2.
\(\frac{\sin x-\sin y}{\sin x \sin y}=\tan \left(\frac{x-y}{2}\right) \cdot \cot \left(\frac{x+y}{2}\right)\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 2

Question 3.
\(\frac{\sin 2 \alpha+\sin 3 \alpha}{\cos 2 \alpha-\cos 3 \alpha}=\cot \left(\frac{\alpha}{2}\right)\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 3

KSEEB Solutions

Question 4.
\(\sin \left(\frac{\pi}{3}+A\right)-\sin \left(\frac{\pi}{3}-A\right)=\sin A\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 4

Question 5.
cos A + cos (120° – A) + cos(120° + A) = 0
Answer:
L.H.S. = cosA + cos A(120° – A) + cos(120° + A)
= cosA + 2cos 120°. cosA = cosA + 2cosA (-cos60°)
= cos A – 2 cos A . \(\frac { 1 }{ 2 }\) = 0 = R.H.S.

Question 6.
sin 65° + cos 65° = \(\sqrt{2} \cos 20^{\circ}\)
Answer:
L.H.S. sin65° + cos65°
= sin(45° + 20°) + cos(45° + 20°)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 5
= 2. \(\frac{1}{\sqrt{2}}\). cos20°= \(\sqrt{2}\) cos20° = R.H.S.

Question 7.
If A + B + C = π, Prove that tan A + tan B + tan C = tan A . tanB · tan C
Answer:
Given A + B + C = π
⇒ A + B = π – C
tan(A + B) = tan(180 – C)
\(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}=-\frac{\tan C}{1}\)
tanA + tanB = – tanc – tanB tanB · tanc
tanA + tanB + tanC = tanA tanB tanC

KSEEB Solutions

Question 8.
Prove that \(\frac{\sin 2 A+\sin 5 A-\sin A}{\cos 2 A+\cos 5 A+\cos A}=\tan 2 A\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 6

Question 9.
If A + B + C = 180°. Prove that cot B. cot C + cot C · cot A + cot A · cot B = 1
Answer:
Given A + B + C = 180°; A + B = 180° – C; cot(A + B) = cot (180° – C)
\(\frac{\cot (\mathrm{A}) \cdot \cot (\mathrm{B})-1}{\cot \mathrm{A}+\cot \mathrm{B}}=-\cot \mathrm{C}\)
cotA . cotB – 1 = -cotA · cotC – cot B . cotC.
∴ cotA . cotB + cotB cotC + cotC . cotA = 1.

Question 10.
If A + B + C = \(\frac{\pi}{2}\) . P.T tanA · tanB + tan B . tan C + tan C · tan A = 1
Answer:
Given A+B+C = \(\frac{\pi}{2}\); A + B = 90° – C; tan(A + B) = tan(90° – C)
\(\frac{\tan A+\tan B}{1-\tan A \tan B}=\cot C=\frac{1}{\tan C}\)
⇒ tanA tanC + tanB tanC = 1 – tanA · tanB
⇒ tan A tanB + tanB tanC + tanC tanA = 1.

Part – C

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 Five Marks Questions and Answers

Prove the following

Question 1
\(\frac{\cos 7 x+\cos 3 x-\cos 5 x-\cos x}{\sin 7 x-\sin 3 x-\sin 5 x+\sin x}=\cot 2 x\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 7

KSEEB Solutions

Question 2.
cos10°.cos30°. cos50°.cos 70° = \(\frac { 3 }{ 16 }\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 8

Question 3.
cos 20° · cos40° · cos60° . cos80° = \(\frac { 1 }{ 16 }\)
Answer:
L.H.S. = cos 20°.cos40°.cos60°.cos80°
= cos60°.cos80°.cos40°.c0s60°
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 9

Question 4.
If A + B + C= 180°C. Prove that sin2a + sin2 \(\left(\frac{2 \pi}{3}+A\right)\) + sin \(\left(\frac{2 \pi}{3}-A\right)\) = \(\frac { 3 }{ 2 }\)
Answer:
Given A + B + C = 180° \(\left[\because \frac{2 \pi}{3}=120^{\circ}\right]\)
sin2A + sin2(120° + A) + sin2(120° – A)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 10

Question 5.
sin 2A + sin 2B + sin 2C = 4 sin A · sinB · sin C
Answer:
L.H.S. = sin2A + sin2B + sin2C
= 2sin(A + B) · cos(A – B) + 2sinC cosC
= 2sinC.cos(A – B) + 2 sinC cosC = 2sinC [cos(A – B) – cos(A + B)]
= 2 sinC[-2 sinA · sin(-B)]
= 4sinA . sinB . sinC ∵ sin(-B)
= R.H.S. = SinB.

KSEEB Solutions

Question 6.
tan 2A + tan 2B + tan 2C = tan 2A . tan 2B . tan 2C
Answer:
Given
A + B + C = 180°
⇒ 2A + 2B + 2C = 360°
2A + 2B = 360° – 2C
tan(2A + 2B) = tan(360° – 2C)
\(\frac{\tan 2 A+\tan 2 B}{1-\tan 2 A \cdot \tan 2 B}=\frac{-\tan 2 C}{1}\)
tan2A + tan2B = – tan2C + tan2A tan2B tan2C
tan2A + tan2B + tan2C = tan 2A tan2B · tan2C

Question 7.
sin 4A + sin 4B + sin 4C = -4sin2A . sin2B · sin2C
Answer:
L.H.S = sin4A + sin4B + sin4C [A+B+C = 180]
= 2 sin(2A + 2B) · cos(2A – 2B) + 2 sinc.cos2C [2A + 2B + 2C = 360]
= (-sin2C) · cos(2A – 2B) + (2 sin2C · cos2C) [2A + 2B = 360 – 20]
= -2 sin 2C [cos(2A – 2B) – cos2C] [sin(2A + 2B) = – sin2C]
= – 2 sin 2C [cos(2A – 2B) – cos(2A + 2B)] [cos(2A + 2B) = cos 2C]
= -2 sin2C[-2 sin2A · sin(-2B)]
= – 4 sin2A · sin2B · sin2C
= R.H.S.

Question 8.
cos2A + cos2B – cos2C = 1 – 2 sin A . sin B · sin C
Answer:
L.H.S. = cos2A + cos2B – cos2C
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 11
= \(\frac { 1 }{ 2 }\)[1+2cos(A+B) cos(A – B) – (2cos2C – 1)]
= \(\frac { 1 }{ 2 }\)[1+2cos(A+B).cos(A – B) – 2cos2C +1]
= \(\frac { 1 }{ 2 }\)[2 + 2(-cosc) cos(A – B) 2cos2C]
= \(\frac { 1 }{ 2 }\)[2 – 2 cos C[cos (A – B) + cos C]
= \(\frac { 1 }{ 2 }\)[2 – 2cosC[cos(A – B) – cos(A+B)]]
= 1 – cosC[-2sin A . sin(-B)]
=1 – cosC[2sin A sin B]
= 1 – 2sinA sinB cosC = R.H.S.

KSEEB Solutions

Question 9.
\(\frac{\sin 2 A+\sin 2 B+\sin 2 C}{\sin 2 A+\sin 2 B-\sin 2 C}=\tan A \cdot \tan B\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 12

Question 10.
cos2A + cos2B + cos2C = -1 – 4 cosA cosB cosC
Answer:
L.H.S. = cos2A + cos2B + cos2C
= 2cos(A + B) · cos(A – B) + 2 cos2C – 1
= -2cosC . cos(A – B) + 2 cos2C – 1
= -2 cosC[cos (A – B) – cos C] – 1
= -2 cosC{cos (A – B) + cos(A + B)} – 1
= -2 cosC (2 cosA · cosB) – 1
= – 1 – 4 cos A cosB cosC = R.H.S.

Question 11.
sin2A + sin2B – sin2C = 2sinA · sin B. cosC
Answer:
L.H.S. = sin2A + sin2B – sin2C
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 13
= \(\frac { 1 }{ 2 }\)[1-(cos 2A + cos2B –cos2C)]
= \(\frac { 1 }{ 2 }\)[1-{{2cos(A+B). Cos(A – B) – 2 cos2C+1}]
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 - 14
= \(\frac { 1 }{ 2 }\)[+2cosC . cos(A – B) +2cos2c]
= \(\frac { 1 }{ 2 }\)[2cosC(cos(A – B)+cosC]
= \(\frac { 1 }{ 2 }\)[2cosC(cos(A – B) – cos(A + B))]
= \(\frac { 1 }{ 2 }\)[2cosC[-2sin A.sin(-B)]]
= 2 sinA sinB sinc

Question 12.
If A + B + C = π. Prove that:
cos 2A + cos 2B – cos2C = 1 – 4 sin A . sin B · cos C
Answer:
L.H.S. = cos2A + cos2B – cos2C
= 2cos(A + B) cos(A – B) – (2 cos2C – 1)
= -2cosC Cos(A – B) – 2 cos2C + 1 [∵ cosC=-cos(A+B)]
= 1 – 2cos C[cos(A – B) – cos (A + B)]
= 1 – 2cos C[-2 sinA . sin(- B)]
= 1 – 4 sin A sin B cos C = R.H.S.

KSEEB Solutions

Question 13.
If A + B + C = 180°. Prove that:
sin 2A – sin 2B + sin 2C = 4 cos A · sin B · cos C
Answer:
Given A + B + C = 180°
2A + 2B + 2C = 360°
2A + 2B = 360° – 2C
sin(2A + 2B) = sin(360° – 20) = – sin2C
cos(2A + 2B) = cos(360° – 2C) = cos 2C
L.H.S = sin2A – sin2B + sin2C
= 2cos(A + B) · sin(A – B) + 2sinC. cosC
= – 2cosC.sin(A – B) + 2 sinc.cosC
= 2 cosC [sinC – sin (A – B)]
= 2 cosC [sin(A + B) – sin(A – B)]
= 2 cos C [2cosA . sinB]
= 4 cosA sinB.cosC = R.H.S.

Question 14.
cos 2A – cos 2B + cos 2C = 1 – 4 sin A · cos B · sin C
Answer:
L.H.S. = cos 2A – cos2B + cos2C
= -2 sin(A + B) . sin(A – B) + 1 – 2sin2C
= 1 – 2 sinC sin (A – B) – 2 sin2C [∵ sin(A+B)-sinc]
= 1 – 2 sinC [sin(A – B) + sinC]
= 1 – 2 sinC [sin(A + B) + sin(A – B)]
= 1 – 2 sinC [2sinA . cosB]
= 1 – 4 sinA cosB sinC = R.H.S.

Leave a Reply

Your email address will not be published. Required fields are marked *