2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Students can Download Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka 2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

2nd PUC Basic Maths Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae One Mark Questions and Answers

Question 1.
Find the value of Sin 70°. Cos 20° + Cos 70°. Sin 20°.
Answer:
= Sin (70° + 20°) = Sin 90° = 1

Question 2.
Write the value of Sin 75°
Answer:
Sin (45° + 30°) = Sin 45. Cos 30 + Cos 45. Sin 30
\(=\frac{\sqrt{3}+1}{2 \sqrt{2}}\)

Question 3.
If Sin \(A=\frac{3}{5}\). Find Sin 2 A
Answer:
Cos \(A=\frac{4}{5}\)
∴ Sin 2 A = 2 Sin A. Cos A
\(=2 \cdot \frac{3}{5} \cdot \frac{4}{5}=\frac{24}{25}\)

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 4.
Write the value of Cos 75° ?
Answer:
Cos (45 + 30°)= Cos 45. Cos 30 – Sin 45. Sin 30
\(=\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

Question 5.
Express the following as sum or difference of two Trigonometric functions.
(i) Sin 5 A. Cos 3 A
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 1

(ii) Cos 55°. Cos 25°
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 2

(iii) \(\cos \frac{5 \theta}{2} \cdot \sin \frac{3 \theta}{2}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 3

(iv) \(\sin \frac{\pi}{5} \cdot \sin \frac{3 \pi}{2}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 4

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 6.
Express each of the following as the product of two trigonometric functions.
(i) Sin 12x + Sin 4x
Answer:
\(=2 \sin \frac{12 x+4 x}{2} \cdot \cos \frac{12 x-4 x}{2}\)
= 2 Sin 8x. Cos 4x.

(ii) Sin 7 A-Sin 3 A
Answer:
\(=2 \cos \frac{7 A+3 A}{2} \cdot \sin \frac{7 A-3 A}{2}\)
= 2 Cos 5 A. Sin 2 A

(iii) Cos 2θ + Cos 4θ
Answer:
\(=2 \cos \frac{2 \theta+4 \theta}{2} \cdot \cos \frac{2 \theta-4 \theta}{2}\)
= 2 Cos 3θ. Cos θ

(iv) Sin 80° – Sin 40°
Answer:
= +2 Cos 120°. Sin 20°.

2nd PUC Basic Maths Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae Two Marks Questions and Answers

Question 1.
P.T Cos (A + B). Cos (A – B) = Cos2 A – Sin2 B.
Answer:
L.H.S = Cos (A + B). Cos (A – B)
= (Cos A. Cos B – sin A. Sin B) (Cos A. Cos B + Sin A Sin B)
= Cos2 A. Cos2 B – Sin2 A. Sin2 B = Cos2 A (1 – Sin2 B) – Sin2 B(1 – Cos2 A)
= Cos2A – Cos2A – Sin2B – Sin2B + Cos2A. – Sin2B
= Cos2 A – Sin2 B = R.H.S.

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 2.
P.T \(\tan \left(\frac{\pi}{4}-A\right)=\frac{1-\tan A}{1+\tan A}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 5

Question 3.
P.T. Cot (A – B) = Cot A. Cot B + 1
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 6

Question 4.
\(\frac{\cos 3 A}{\sec A}+\frac{\sin 3 A}{\csc A}=\cos _{2} A\)
Answer:
L.H.S = Cos 3 A. Cos A + Sin 3 A. Sin A
= Cos (3 A – A)
= Cos2 A = R.H.S.

Question 5.
P. T. tan 75° + Cot 75° = 4
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 7

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 6.
P. T. Sin 105° + Cos 105° = \(\frac{1}{\sqrt{2}}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 8
Question 7.
If Sin A \(\frac{3}{5}\), Cos B = \(\frac{4}{5}\) find Sin (A+B) ,Cos(A -B)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 9

Question 8.
P.T \(\cos \left(A+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\) (CosA – SinA)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 10

Question 9.
If tan \(\alpha=\frac{n}{n+1}\) and tan β \(=\frac{1}{2 n+1}\)
S.T. α +β = \(\frac{\pi}{4}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 11

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 10.
P.T. \(\cos \left(\frac{\pi}{6}+\mathbf{A}\right) \cdot \cos \left(\frac{\pi}{6}-\mathbf{A}\right)-\sin \left(\frac{\pi}{6}+\mathbf{A}\right) \cdot \sin \left(\frac{\pi}{6}-\mathbf{A}\right)=\frac{\sqrt{3}}{2}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 12

Question 11.
P. T. Cot 2 A + Tan A = Cosec 2 A
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 13

Question 12.
P.T. Sin 2A = 2 Sin A. Cos A
Answer:
W.K.T. Sin (A + B) = Sin A. Cos B +Cos A Sin B
Put  B = A
Sin (A + A) = Sin A Cos A + Cos A Sin A
Sin 2A = 2 Sin A. Cos A.

Question 13.
P.T. Cos 2 A = Cos2 A – Sin2 A = 2 Cos2 A – 1 = 1 – 2 Sin2 A.
Answer:
(i) W.K.T Cos (A + B) = Cos A. Cos B – Sin A Sin B
Put  B = A
Cos (A + A) = cos A. Cos A – Sin A Sin A
∴ Cos 2 A = Cos2 A – Sin2 A  ………………………… (1)

(ii) W.K.T. Cos2A = Cos2 A – Sin2 A
Cos2 A = Cos2 A – (1 – Cos2 A)
∵  Sin2 A = I – Cos2 A
= Cos2 A – 1 + Cos2 A
= 2 Cos2 A – 1
∴ Cos2 A= 2 Cos2 A – 1 ………………………… (2)

(iii) W.K.T. Cos2 A = Cos2 A – Sin2 A
= 1 – Sin2 A – Sin2 A
∵ Cos2 A = 1 – Sin2 A
Cos2 A = 1 – 2 Sin2 A  ………………………… (3)

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 14.
P.T. \(Tan 2A = \frac{2 \tan A}{1-\tan ^{2} A}\frac{2 \tan A}{1-\tan ^{2} A}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 14

Question 15.
If Cos A = \(\frac{4}{5}\)find Cos 3 A
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 15

Question 16.
P.T \(\frac{\cos 2 A}{1+\sin 2 A}=\frac{\cos A-\sin A}{\cos A+\sin A}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 16

Question 17.
P.T \(\frac{\sin A+\sin 2 A}{1+\cos A+\cos 2 A}=\tan A\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 17

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 18.
S.T \( Tan A = \tan A=\frac{\tan (A-B)+\tan B}{1-\tan (A-B) \cdot \tan B}\)
Answer:
R.H.S – tan (A – B + B) = tan A = R.H.S
Using tan (A + B) formula. Where A = (A – B).

Question 19.
P.T \(\frac{1+\sin 2 \theta}{\cos 2 \theta}=\frac{1+\tan \theta}{1-\tan \theta}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 19

Question 20.
P.T. (Cos4 θ – Sin4 θ) = 2 Cos2 θ -1.
Answer:
L.H.S = Cos4 θ – Sin4 θ
= (Cos2 θ)2 – (Sin2 θ)2 = (Cos2θ – Sin2θ) (Cos2 θ + Sin2 θ)
= (Cos2 θ – Sin2 θ) x 1 = Cos2 θ – (1 – Cos2 θ)
= Cos2 θ – 1 + Cos2 θ = 2 Cos2 θ – 1 = R.H.S.

Question 21.
P.T \(\frac{\cos ^{3} A-\sin ^{3} A}{\cos A-\sin A}=1+\frac{1}{2} \sin 2 A\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 20
Question 22.
P.T \(\frac{\cos 2 A-\cos 12 A}{\sin 12 A-\sin 2 A}=\tan 7 A\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 21

Question 23.
P.T \(\frac{\sin x-\sin y}{\sin x+\sin y}=\tan \left(\frac{x-y}{2}\right) \cot \left(\frac{x+y}{2}\right)\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 22

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 24.
If A + B + C = π P.T. tan A + tan B + tan C = tan A. tan B. tan C.
Answer:
Given A + B + C = π
A + B = π – C
tan (A + B) = tan (π – C)
\(\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}=\frac{-\tan C}{1}\)
tan A + tan B = – tan C (1 – tan A tan B)
= – tan C + tan A. tan B. tan C.
tan A + tan B + tan C = tan A tan B tan C.

Question 25.
P.T \(\frac{\sin 2 A+\sin 5 A-\sin A}{\cos 2 A+\cos 5 A+\cos A}=\tan ^{2} A\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 23

Question 26.
P.T. Sin 65 + Cos 65 = \(\sqrt{2} \)Cos 20°.
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 56

Question 27.
If A + B + C = \(\frac{\pi}{2}\) P.T. tan A. tan B + tan B . tan C + tan C tan A = 1
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 24

Question 28.
P.T \(\sin \left(\frac{\pi}{3}+\mathbf{A}\right)-\sin \left(\frac{\pi}{3}-\mathbf{A}\right)=\sin \mathbf{A}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 25

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 29.
P.T. Cos A + Cos (120 – A) + Cos (120 + A) + Cos (120 + A) = 0
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 26

Question 30.
P.T \(\frac{\sin 2 \alpha+\sin 3 \alpha}{\cos 2 \alpha-\cos 3 \alpha}=\cot \left(\frac{\alpha}{2}\right)\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 27

2nd PUC Basic Maths Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae Five Marks Questions and Answers

Question 1.
P. T Sin 3θ = 3 Sin θ – 4 Sin3 θ
Answer:
L.H.S. = Sin3 θ = Sin (θ + 20)
= Sin θ. Cos 2θ + Cos θ Sin 2θ
= Sin θ (1 – 2 Sin2 θ) + Cos θ. (2 Sin θ. Cos θ)
= Sin θ – 2 Sin3 θ + 2 Sin θ. Cos2 θ
= Sin θ – 2 Sin3 θ + 2 Sin θ (1 – Sin2 θ)
= Sin θ – 2 Sin3 θ + 2 Sinθ -2 Sin3 θ
= 3 Sin θ – 4 Sin
∴ Sin3θ = 3Sinθ – 4Sin3θ

Question 2.
P. T. Cos 3θ = 4 Cos3 θ – 3 Cos θ.
Answer:
L.H.S. Cos 3θ = Cos (θ + 2θ)
= Cos θ. Cos 2θ – Sin θ. Sin 2 θ
= Cos θ (2 Cos2 θ – 1) – Sin θ. 2 Sin θ . Cos θ
= 2 Cos3 θ – Cos θ – 2 Cos θ Sin2 θ
= 2 Cos3 θ – Cos θ – 2 Cos θ (1 – Cos2 θ)
= 2 Cos3 θ – Cos θ – 2 Cos θ + 2 Cos3 θ
= 4 Cos3 θ – 3 Cos θ
∴ Cos3θ = 4Cos3 θ- 3Cosθ .

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 3.
P.T \(\tan 3 \theta=\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 28
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 29

Question 4.
P.T \(\frac{\sin 3 \theta}{\sin \theta}-\frac{\cos 3 \theta}{\cos \theta}=2\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 30

Question 5.
P.T \(\frac{\cos 3 A}{2 \cos 2 A-1}=\cos A\) and Hence find cos 15°
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 31

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 6.
P.T  \(\frac{1+\cos 2 A+\sin 2 A}{1-\cos 2 A+\sin 2 A}=C o t A\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 32

Question 7.
P.T. Cos6A + Sin6A = 1- \(\frac{3}{4}\) Sin2 (2A)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 33

Question 8.
It tan α \(\frac{1}{3}\) , tan β = \(\frac{1}{7}\) P.T tan (2α + β) – 45°
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 34
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 35

Question 9.
If \(\sin A=\frac{5}{13} \cos B=\frac{-4}{5}\), A is in IInd Quadrant and B is in IIIrd Quadrant. Find (i) Sin (A + B) (ii) Cos (A + B) (iii) tan (A + B), (iv) Sin (A – B)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 36

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 10.
P.T. Cos (120°+A) + Cos (120°-A) = -Cos A.
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 37

Question 11.
P.T\(\frac{\sin (A+B)}{\sin (A-B)}=\frac{\tan A+\tan B}{\tan A-\tan B}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 38

Question 12.
P.T. tan 2θ – tan θ = tan θ. Sec 2θ.
Sol.
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 39

Question 13.
P.T \(\frac{\cos 7 x+\cos 3 x-\cos 5 x+\cos x}{\sin 7 x-\sin 3 x-\sin 5 x+\sin x}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 40
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 41

Question 14.
If A + B + C = 180° P.T. Sin 2A + Sin 2 B + Sin 2 C = 4 Sin A. Sin B Sin C.
Answer:
GivenA + B + C= 180
L.H.S. = Sin 2 A + Sin 2 B + Sin 2 C
\(=2 \sin \frac{2 A+2 B}{2} \cdot \cos \frac{2 A-2 B}{2}+2 \sin C \cdot \cos C\)
∵ Sin 2 C = 2 Sin C. Cos C
= 2 Sin (A + B). Cos (A – B) + 2 Sin (C). Cos C
= 2 Sin C. Cos (A – B) + 2 Sin C. Cos C                                 Sin (A + B) = Sin C
= 2 sin C [Cos (A – B) + Cos C]                                             Cos(A + B) = -CosC
= 2 Sin C [Cos (A – B) – Cos (A + B)]
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 42
= -4 Sin C. Sin A. -Sin B
= 4 Sin A. Sin B. Sin C = R.H.S.

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 15.
P.T. Cos 2 A+ Cos 2 B + Cos 2 C = – 1 – 4 Cos A. Cos B. Cos C.
Answer:
L.H.S. = Cos 2 A + Cos 2 B + Cos 2 C
\(=2 \cos \frac{2 A+2 B}{2} \cdot \cos \frac{2 A-2 B}{2}+2 \cos ^{2} C-1\)
= 2 Cos (A + B). Cos (A – B) + 2 Cos2C – 1    ∵ Cos (A + B) = – 2 Cos C
= – 1 – 2Cos CCos (A + B) + 2 Cos2 C
= – 1 – 2 Cos C [cos (A + B) – Cos C]
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 43
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 44

Question 16.
P. T. Sin2 A + Sin2 B – Sin2 C = 1 – 4 Sin A. Sin B. Cos C.
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 45

Question 17.
P.T. Cos2 A + Cos2 B – Cos2 C = 1 – 2 Sin A Sin B Cos C.
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 46
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 47
= 1-Sin A [Sin (B + C) + sin (B – C)
= 1 – Sin A (2 Sin B. Cos C)
= 1 – 2 Sin A. Sin B. Sin C.

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 18.
P.T. tan2 A + tan 2 B + tan 2 C = tan 2A. tan 2 B. tan 2 C.
Answer:
Given A + B + C = 180°
2A + 2B + 2C = 360°
2A + 2B = 360 -2C
Taking tan both sides
tan (2 A + 2B) = tan (360 – 2C)
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 57
tan2A + tan2B = – tan2C (1 – tan 2 A tan 2B)
= – tan 2 A + tan 2 B = – tan 2 C
= – tan 2C + tan 2 A. tan 2 B . tan 2 C

Question 19.
S.T \(\Sigma \frac{\sin (\mathbf{A}-\mathbf{B})}{\cos \mathbf{A} \cdot \mathbf{C} \mathbf{o s} \mathbf{B}}=\mathbf{0}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 48
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 49

Question 20.
P.T tan A. tan 3 A. tan 4 A = tan 4 A – tan 3 A – tan A
Answer:
Consider tan 4 A = tan (A + 3 A).
\(\frac{\tan 4 A}{1}=\frac{\tan A+\tan 3 A}{1-\tan A \cdot \tan 3 A}\)
tan A +tan 3 A = tan 4 A (1 – tan A. tan 3A)
tan A + tan 3 A = tan 4 A – tan A. tan 3 A . tan 4 A
∴ tan A. tan 3 A. tan 4 A = tan 4 A-tan 3 A – tan A.

Question 21.
T. Sin 105° + Cos 105° = 2 Cos 15°.
Answer:
L.H.S. = Sin 105° + Cos (180 – 75°)
= Sin 105° + Sin 75°
= 2 sin \(\left(\frac{105+75}{2}\right) \cos \left(\frac{105-75}{2}\right)\)
= 2 Sin 90°.Cos 15°
= 2.1.Cos 15°
= 2 Cos 15° = R.H.S

2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae

Question 22.
P.T \(\cos \left(\frac{\pi}{4}-A\right)-\sin \left(\frac{\pi}{4}+A\right)=0\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 50

Question 23.
P.T \(\sin \left(\frac{\pi}{3}-A\right) \cdot \cos \left(\frac{\pi}{6}+A\right)+\cos \left(\frac{\pi}{3}-A\right) \cdot \sin \left(\frac{\pi}{6}+A\right)=1\)
Answer:
L.H.S. = Sin A. Cos B + Cos A. Sin B
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 51

Question 24.
P. T. Cos 20°. Cos 40°. Cos 80° = \(\frac{1}{8}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 52

Question 25.
P.T. Cos 10° . Cos 30° . Cos 50° Cos 70° \(=\frac{3}{16}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 53
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 54

Question 26.
P.T. Cos2 A + Cos2 (60 + A) + Cos2 (60 – A) \(=\frac{3}{16}\)
Answer:
2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Sub Multiples Angles and Transformation Formulae 55

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