2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.2

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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 – 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
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 sin³10°
Answer:
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
Answer:
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 = \(\frac{\tan (A-B)+\tan B}{1-\tan (A-B) \tan B}\)
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
= sin²A + cos²A – 2sinA cosA
= 1 – sin2A = R.H.S.

Question 15.
Prove that cos⁴θ – sin⁴θ = 2 cos²θ – 1
Answer:
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 \(\frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A}=1+\frac{1}{2} \sin 2 A\)
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.
cos⁶A+ sin⁶A = 1 – \(\frac { 3 }{ 4 }\) sin³(2A)
Answer:
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 \(\frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta}=2\)
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 tan²(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: