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

Answer:

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

Answer:

Question 3.

If tan A = \(\frac{1}{\sqrt{3}}\) find tan 2A

Answer:

Question 4.

If sin A = \(\frac { 3 }{ 5 }\) find sin 3A

Answer:

Given sin A = \(\frac { 3 }{ 5 }\) sin3A = 3sinA – 4sin^{3}A = \(3 \cdot \frac{3}{5}-4\left(\frac{3}{5}\right)^{3}\)

\(=\frac{9}{5}-4 \cdot \frac{27}{125}=\frac{225-108}{125}=\frac{117}{125}\)

Question 5.

If cos A = \(\frac { 4 }{ 5 }\) 4 find cos 3A

Answer:

Given cos A = \(\frac { 4 }{ 5 }\)

Question 6.

If tan A = \(\frac { 3 }{ 4 }\) find tan 3A

Answer:

Question 7.

Find the value of 3 sin 10° – 4 sin^{3}10°

Answer:

3sin10° – 4sin^{3}10° is of the form 3 sin A – 4sin^{3}A = sin3A = sin 3-10° = sin 30° = \(\frac { 1 }{ 2 }\)

Question 8.

If cot A = \(\frac { 12 }{ 5 }\) and A is acute find sin3A

Answer:

Given cot A = \(\frac { 12 }{ 5 }\) ⇒ sin A = \(\frac { 5 }{ 13 }\) cosA = \(\frac { 12 }{ 13 }\)

∴ sin 3A = 3sinA – 4 sin^{3}A

Question 9.

Show that tan A = \(\frac{\tan (A-B)+\tan B}{1-\tan (A-B) \tan B}\)

Answer:

Question 10.

Prove that \(\frac{\cos 2 A}{1+\sin 2 A}=\frac{\cos A-\sin A}{\cos A+\sin A}\)

Answer:

Question 11.

Prove that \(\frac{1+\sin 2 \theta}{\cos 2 \theta}=\frac{1+\tan \theta}{1-\tan \theta}\)

Answer:

Question 12.

Prove that \(\frac{\sin A+\sin 2 A}{1+\cos A+\cos 2 A}=\tan A\)

Answer:

Question 13.

Prove that \(\frac{\sin 2 \theta}{1+\cos 2 \theta}=\tan \theta\)

Answer:

Question 14.

Prove that (sin A – cos A)^{2} = 1 – sin 2A

Answer:

L.H.S (sinA – cosA)^{2}

= sin^{2}A + cos^{2}A – 2sinA cosA

= 1 – sin2A = R.H.S.

Question 15.

Prove that cos^{4}θ – sin^{4}θ = 2 cos^{2}θ – 1

Answer:

L.H.S. cos^{4}θ – sin^{4}θ

= (cos^{2}θ)^{2} – (sin^{2}θ)^{2}

= (cos^{2}θ + sin^{2}θ) (cos^{2}θ – sin^{2}θ)

= 1 . (cos^{2}θ – (1 – cos^{2}θ)

= cos^{2}θ – 1 + cos^{2}θ = 2cos^{2}θ – 1 = R.H.S.

Question 16.

Prove that \(\frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A}=1+\frac{1}{2} \sin 2 A\)

Answer:

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.

\(\frac{1+\cos 2 A+\sin 2 A}{1-\cos 2 A+\sin 2 A}=\cot A\)

Answer:

Question 2.

\(\frac{\cos 3 A}{2 \cos 2 A-1}=\cos A\) and hence find cos 15°

Answer:

Question 3.

\(\frac{1-\cos 2 A+\sin 2 A}{1+\cos 2 A+\sin 2 A}=\tan A\)

Answer:

Question 4.

cos^{6}A+ sin^{6}A = 1 – \(\frac { 3 }{ 4 }\) sin^{3}(2A)

Answer:

cos^{6}A+ sin^{6}A [∵ a^{3} + b^{3} = (a + b)^{3} – 3ab(a+b)]

= (cos^{2}A)^{3} + (sin^{2}A)^{3}

= (cos^{2}A + sin^{2}A)^{3} – 3cos^{2}A – sin^{2}A(cos^{3}A + sin^{3}A)

= 1^{3} – 3cos^{2}A · sin^{2}A.1

Question 5.

Provfe that \(\frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta}=2\)

Answer:

Question 6.

sec (45° + A) · sec(45° – A) = 2 sec 2A

Answer:

Question 7.

\(\frac{\cot A}{\cot A-\cot 3 A}+\frac{\tan A}{\tan A-\tan 3 A}=1\)

Answer:

Question 8.

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

Answer:

Question 9.

If tan α = \(\frac { 1 }{ 3 }\), tan β = \(\frac { 1 }{ 7 }\) P.T. tan(2α + β) = 45°.

Answer:

L.H.S = tan(2α + β)

tan(2α + β) = 1 = tan 45° = R.H.S.

Question 10.

If tan^{2}(45° + θ) = \(\frac { a }{ b }\) prove that \(\frac{b-a}{b+a}\) = – sin 2θ

Answer:

Question 11.

Prove that tan 2θ – tanθ = tan θ . sec 2θ.

Answer:

L.H.S = tan2θ – tanθ

Question 12.

Prove that cos 2α – tan α = \(\frac{\cos 3 \alpha}{\cos \alpha \cdot \sin 2 \alpha}\)

Answer: