Students can Download Basic Maths Exercise 14.2 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 14.2

Part – A & B

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.2 One or Two Marks Questions and Answers

Question 1.
If sin A = $$\frac { 1 }{ 2 }$$ find sin 2A
Given sin A = $$\frac { 1 }{ 2 }$$ ⇒ cos A = $$\frac{\sqrt{3}}{2}$$ sin2A

Question 2.
If cos A = $$\frac{\sqrt{3}}{2}$$ find cos 2A

Question 3.
If tan A = $$\frac{1}{\sqrt{3}}$$ find tan 2A

Question 4.
If sin A = $$\frac { 3 }{ 5 }$$ find sin 3A
Given sin A = $$\frac { 3 }{ 5 }$$ sin3A = 3sinA – 4sin3A = $$3 \cdot \frac{3}{5}-4\left(\frac{3}{5}\right)^{3}$$
$$=\frac{9}{5}-4 \cdot \frac{27}{125}=\frac{225-108}{125}=\frac{117}{125}$$

Question 5.
If cos A = $$\frac { 4 }{ 5 }$$ 4 find cos 3A
Given cos A = $$\frac { 4 }{ 5 }$$

Question 6.
If tan A = $$\frac { 3 }{ 4 }$$ find tan 3A

Question 7.
Find the value of 3 sin 10° – 4 sin310°
3sin10° – 4sin310° is of the form 3 sin A – 4sin3A = sin3A = sin 3-10° = sin 30° = $$\frac { 1 }{ 2 }$$

Question 8.
If cot A = $$\frac { 12 }{ 5 }$$ and A is acute find sin3A
Given cot A = $$\frac { 12 }{ 5 }$$ ⇒ sin A = $$\frac { 5 }{ 13 }$$ cosA = $$\frac { 12 }{ 13 }$$
∴ sin 3A = 3sinA – 4 sin3A

Question 9.
Show that tan A = $$\frac{\tan (A-B)+\tan B}{1-\tan (A-B) \tan B}$$

Question 10.
Prove that $$\frac{\cos 2 A}{1+\sin 2 A}=\frac{\cos A-\sin A}{\cos A+\sin A}$$

Question 11.
Prove that $$\frac{1+\sin 2 \theta}{\cos 2 \theta}=\frac{1+\tan \theta}{1-\tan \theta}$$

Question 12.
Prove that $$\frac{\sin A+\sin 2 A}{1+\cos A+\cos 2 A}=\tan A$$

Question 13.
Prove that $$\frac{\sin 2 \theta}{1+\cos 2 \theta}=\tan \theta$$

Question 14.
Prove that (sin A – cos A)2 = 1 – sin 2A
L.H.S (sinA – cosA)2
= sin2A + cos2A – 2sinA cosA
= 1 – sin2A = R.H.S.

Question 15.
Prove that cos4θ – sin4θ = 2 cos2θ – 1
L.H.S. cos4θ – sin4θ
= (cos2θ)2 – (sin2θ)2
= (cos2θ + sin2θ) (cos2θ – sin2θ)
= 1 . (cos2θ – (1 – cos2θ)
= cos2θ – 1 + cos2θ = 2cos2θ – 1 = R.H.S.

Question 16.
Prove that $$\frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A}=1+\frac{1}{2} \sin 2 A$$

Part – C

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.2Five Marks Question and Answers

Prove the following

Question 1.
$$\frac{1+\cos 2 A+\sin 2 A}{1-\cos 2 A+\sin 2 A}=\cot A$$

Question 2.
$$\frac{\cos 3 A}{2 \cos 2 A-1}=\cos A$$ and hence find cos 15°

Question 3.
$$\frac{1-\cos 2 A+\sin 2 A}{1+\cos 2 A+\sin 2 A}=\tan A$$

Question 4.
cos6A+ sin6A = 1 – $$\frac { 3 }{ 4 }$$ sin3(2A)
cos6A+ sin6A [∵ a3 + b3 = (a + b)3 – 3ab(a+b)]
= (cos2A)3 + (sin2A)3
= (cos2A + sin2A)3 – 3cos2A – sin2A(cos3A + sin3A)
= 13 – 3cos2A · sin2A.1

Question 5.
Provfe that $$\frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta}=2$$

Question 6.
sec (45° + A) · sec(45° – A) = 2 sec 2A

Question 7.
$$\frac{\cot A}{\cot A-\cot 3 A}+\frac{\tan A}{\tan A-\tan 3 A}=1$$

Question 8.
Prove that $$\frac{\cos 2 A}{1+\sin 2 A}$$ = tan(45° – A)

Question 9.
If tan α = $$\frac { 1 }{ 3 }$$, tan β = $$\frac { 1 }{ 7 }$$ P.T. tan(2α + β) = 45°.
L.H.S = tan(2α + β)

tan(2α + β) = 1 = tan 45° = R.H.S.

Question 10.
If tan2(45° + θ) = $$\frac { a }{ b }$$ prove that $$\frac{b-a}{b+a}$$ = – sin 2θ
Prove that cos 2α – tan α = $$\frac{\cos 3 \alpha}{\cos \alpha \cdot \sin 2 \alpha}$$