Students can Download Basic Maths Exercise 14.1 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 14.1

Part – A

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 Two Marks Questions and Answers

Question 1.
Obtain the values of the trigonometric functions of 75°, 15° and 105° and prove that
(i) tan 75° + cot 75° = 4
(ii) sin 105° + cos 105° = \(\frac{1}{\sqrt{2}}\)
(iii) Sec 15° + cosec 15° = \(2 \sqrt{6}\)
Answer:
sin(45° + 30°) = sin45°. cos30° + cos45° · sin 30° = \(\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\)
∴ sin 75° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
sin15° = sin(45° – 30°) = \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
sin(105°) = sin(60° + 45°) = sin60°. sin45° + cos60°.cos45°
= \(\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}}+\frac{1}{2} \frac{1}{\sqrt{2}}\) ; sin 105° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
cos(75°) = cos(45° + 30°) = cos 45° – cos30° + sin45° . sin30° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
cos105° = cos(60° + 45°) = cos60°. cos45° – sin60° . sin45° = \(\frac{+1-\sqrt{3}}{2 \sqrt{2}}\)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 1

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Question 2.
If sin A = \(\frac { 3 }{ 5 }\), cos B = \(\frac { 4 }{ 5 }\) find sin(A + B) and cos(A – B). Where A and B are acute angles.
Answer:
Given sin A = \(\frac { 3 }{ 5 }\) , cos B = \(\frac { 4 }{ 5 }\)
⇒ cos A = \(\frac { 4 }{ 5 }\) & sin B = \(\frac { 3 }{ 5 }\) ∵ sin2A + cos2A = 1
(i) sin(A + B) = sinA cosB + cosA sinB = \(\frac{3}{5} \cdot \frac{4}{5}+\frac{4}{5} \cdot \frac{3}{5}=\frac{24}{25}\)
(ii) cos(A – B) = COSA COSB + sinA sinB = \(\frac{4}{5} \cdot \frac{4}{5}+\frac{3}{5} \frac{3}{5}=\frac{16}{25}+\frac{9}{25}=\frac{25}{25}=1\)

Question 3.
If cos A = \(\frac { 5 }{ 13 }\), cos B = \(\frac { 24 }{ 25 }\) find cos(A + B) and sin (A – B). A and B are acute angles.
Answer:
Given
cos A = \(\frac { 5 }{ 13 }\) cosB = \(\frac { 24 }{ 25 }\)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 2
cos (A + B) = cosA cosB – sinA sinB = \(\frac{5}{13} \cdot \frac{24}{25}-\frac{12}{13} \cdot \frac{7}{25}=\frac{120-84}{325}=\frac{36}{325}\)
sin(A – B) = sinA . cosB – cosA sinB = \(\frac{12}{13} \cdot \frac{24}{25}-\frac{5}{13} \cdot \frac{7}{25}=\frac{288-35}{325}=\frac{253}{325}\)

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Question 4.
If sec A = \(\frac { 17 }{ 8 }\) , cosec B = \(\frac { 5 }{ 4 }\) find set (A + B). cosec(A + B), A and B are acute angles
Answer:
Given sec = \(\frac { 17 }{ 8 }\) ⇒ cos A = \(\frac { 8 }{ 17 }\) ∴ sin A = \(\frac { 15 }{ 17 }\) cosec B = \(\frac { 5 }{ 4 }\), sinB = \(\frac { 4 }{ 5 }\), cosB = \(\frac { 3 }{ 5 }\)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 3
sin(A – B) = sinA cosB – cosA sinB
\(=\frac{15}{17} \cdot \frac{3}{5}-\frac{8}{17} \cdot \frac{4}{5}=\frac{45}{85}-\frac{32}{85}=\frac{13}{85}\) ; cosec (A – B) = \(\frac { 85 }{ 13 }\)

Question 5.
If sin A = \(\frac { 3 }{ 5 }\) cos B = \(\frac { -8 }{ 17 }\) < A < π and \(\frac{\pi}{2}\) < B < π. Find the values of sin(A + B) and cos(A – B).
Answer:
Given sin A = \(\frac { 3 }{ 5 }\) . cos A = \(\frac { -4 }{ 5 }\)
cos B = \(\frac { -8 }{ 17 }\), sin B = \(\frac { 15 }{ 17 }\)
sin(A + B) = sinA cosB + Cos A sinB = \(\frac{3}{5}\left(\frac{-8}{17}\right)+\left(\frac{-4}{5}\right) \frac{15}{17}=\frac{-24-60}{85}=\frac{-84}{85}\)

KSEEB Solutions

Question 6.
If sin A = \(\frac { 7 }{ 25 }\), cos B = \(\frac { -12 }{ 13 }\) where \(\frac{\pi}{2}\) < A < π and π < B <  \(\frac{3 \pi}{2}\) find the values of:
(i) sin(A + B)
(ii) cos(A + B)
(iii) sin(A – B)
(ii) cos(A – B)
(v) tan(A + B)
(vi) tan(A – B)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 4
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 5

KSEEB Solutions

Question 7.
If tan A = \(\frac { 1 }{ 2 }\), tan B = \(\frac { 1 }{ 3 }\) Find tan(A + B), tan(A – B)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 6

Question 8.
tan(A – B) = \(\frac { 1 }{ 7 }\), tan A = \(\frac { 1 }{ 2 }\) show that A + B = 45°.
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 7

KSEEB Solutions

Question 9.
If tan A = \(\frac { 1 }{ 3 }\) tan (A + B) = \(\frac { 1 }{ 7 }\) find tan B.
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 8

Question 10.
tan A = \(\frac { 3 }{ 4 }\) and tan B = \(\frac { 1 }{ 7 }\) . show that tan (A + B) = 1.
Answer:
Given tan A = \(\frac { 3 }{ 4 }\), tanB = \(\frac { 1 }{ 7 }\)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 9

Question 11.
If tan A = \(\frac { 5 }{ 6 }\) and tan(A + B) = 1 Show that tanB = \(\frac { 1 }{ 11 }\)
Answer:
Given
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 10

KSEEB Solutions

Question 12.
If tan α = \(\frac{\mathrm{n}}{\mathrm{n}+1}\) and tan β = \(\frac{1}{2 n+1}\) Show that α + β = \(\frac{\pi}{4}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 11

Question 13.
Prove that \(\frac{\cos 2 A}{\sec A}+\frac{\sin 2 A}{\csc A}=\cos A\)
Answer:
L.H.S. = cos2A . \(\frac{1}{\sec A}\) + sin 2A . \(\frac{1}{\csc A}\)
= cos2A . cosA + sin2A . sinA
= cos(2A – A) = cosA = R.H.S.

Question 14.
Prove that sin(45° + A) + cos(45° + A) = \(\sqrt{2}\) cosA ,
Answer:
L.H.S. = sin(45° + A) + cos(45° + A)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 12

KSEEB Solutions

Question 15.
Prove that cos \(\left(A+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\) (cos A – sin A).
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 27

Question 16.
Prove that cos \(\left(\frac{\pi}{6}+\mathbf{A}\right)\) . cos \(\left(\frac{\pi}{6}-\mathbf{A}\right)\) – sin\(\left(\frac{\pi}{6}+\mathbf{A}\right)\) . sin \(\left(\frac{\pi}{6}+\mathbf{A}\right)\) = \(\frac { 1 }{ 2 }\).
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 13

KSEEB Solutions

Part – B

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 Five Marks Questions and Answers

Question 1.
Prove the following \(\frac{\sin (A+B)}{\cos A \cdot \cos B}\) =tan A + tanB.
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 14

Question 2.
\(\frac{\sin (A-B)}{\sin A \sin B}=\cot B-\cot A\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 15

Question 3.
Show that cos(A + B) cos(A – B) = cos2A – sin2B = cos2B – sin2A.
Answer:
L.H.S = cos(A + B).cos(A – B)
= (cos A . cosB – sin A sinB) (cos A cosB + sinA sinB)
= cos2A . cos2B – sin2A · sin 2B = cos2A(1 – sin2B) – (1 – cos2A) · sin2B
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 16
Similarly cos2A . cos2B – sin2A · sin2B = (1 – sin2A) cos2B – sin2A(1 – cos2B)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 17

KSEEB Solutions

Question 4.
sin(A + B) sin(A – B) = cos2B – Cos2A
Answer:
L.H.S = sin(A + B) sin(A – B) = (sinA cosB + cosA sinB) (sin A cosB – CosA sinB)
= sin2A cos2B – cos2A sin2B = (1 – Cos2A) cos2B – cos2A (1 – cos2B)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 18

Question 5.
\(\cos \left(\frac{\pi}{4}-A\right)-\sin \left(\frac{\pi}{4}+A\right)=0\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 19

Question 6.
\(\sin \left(\frac{\pi}{3}-A\right) \cdot \cos \left(\frac{\pi}{6}+A\right)+\cos \left(\frac{\pi}{3}-A\right) \cdot \sin \left(\frac{\pi}{6}+A\right)=1\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 20

KSEEB Solutions

Question 7.
tan 15° + cot 15° = 4.
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 21

Question 8.
sin 105° + cos 105° = cos 45°
Answer:
L.H.S = sin105° + cos(105°) = sin(60° + 45°) + cos(60° + 459)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 22

Question 9.
cot 2A + tan A = cosec 2A.
Answer:
L.H.S. = cot2A + tanA
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 23

KSEEB Solutions

Question 10.
P.T tan A tan 3A, tan 4A = tan 4A – tan3A – tanA.
Answer:
Consider tan4A = tan(3A + A)
\(\frac{\tan 4 A}{1}=\frac{\tan 3 A+\tan A}{1-\tan 3 A \cdot \tan A}\) ; tan4A(1 – tan3A · tanA) = tan3A + tanA
tan4A – tan4A tan3A tanA = tan3A + tanA
tan4A – tan3A – tanA = tan4A · tan3A tanA

Question 11.
Prove that cos(45° – A) · cos(45° – B) – sin(45° – A) sin(45° – B) = sin(A + B).
Answer:
L.H.S = cos(45° – A) · cos(45° – B) – sin(45° – A) · sin(45° – B)
= cos(45° – A + 45° – B) ∵ cosA . cosB – sinA . sinB
= cos(90° – (A + B)) = cos (A + B) = sin(A + B) = R.H.S.

Question 12.
Show that \(\sum \frac{\sin (A-B)}{\cos A \cdot \cos B}=0\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 24

Question 13.
cos(120° + A) + cos(120° – A) = -cos A
Answer:
L.H.S = cos(120° + A) + cos(120° – A)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 25

KSEEB Solutions

Question 14.
\(\frac{\sin (A+B)}{\sin (A-B)} \quad \frac{\tan A+\tan B}{\tan A-\tan B}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 - 26

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