2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1

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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° = \(\frac{+1-\sqrt{3}}{2 \sqrt{2}}\)

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 = \(\frac { 24 }{ 25 }\)

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 = \(\frac{12}{13} \cdot \frac{24}{25}-\frac{5}{13} \cdot \frac{7}{25}=\frac{288-35}{325}=\frac{253}{325}\)

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 = \(\frac { 3 }{ 5 }\)

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 = \(\frac{3}{5}\left(\frac{-8}{17}\right)+\left(\frac{-4}{5}\right) \frac{15}{17}=\frac{-24-60}{85}=\frac{-84}{85}\)

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:

Question 7.
If tan A = \(\frac { 1 }{ 2 }\), tan B = \(\frac { 1 }{ 3 }\) Find tan(A + B), tan(A – B)
Answer:

Question 8.
tan(A – B) = \(\frac { 1 }{ 7 }\), tan A = \(\frac { 1 }{ 2 }\) show that A + B = 45°.
Answer:

Question 9.
If tan A = \(\frac { 1 }{ 3 }\) tan (A + B) = \(\frac { 1 }{ 7 }\) find tan B.
Answer:

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 = \(\frac { 1 }{ 7 }\)

Question 11.
If tan A = \(\frac { 5 }{ 6 }\) and tan(A + B) = 1 Show that tanB = \(\frac { 1 }{ 11 }\)
Answer:
Given

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:

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)

Question 15.
Prove that cos \(\left(A+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\) (cos A – sin A).
Answer:

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:

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:

Question 2.
\(\frac{\sin (A-B)}{\sin A \sin B}=\cot B-\cot A\)
Answer:

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

Similarly cos2A . cos2B – sin2A · sin2B = (1 – sin2A) cos2B – sin2A(1 – cos2B)

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)

Question 5.
\(\cos \left(\frac{\pi}{4}-A\right)-\sin \left(\frac{\pi}{4}+A\right)=0\)
Answer:

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:

Question 7.
tan 15° + cot 15° = 4.
Answer:

Question 8.
sin 105° + cos 105° = cos 45°
Answer:
L.H.S = sin105° + cos(105°) = sin(60° + 45°) + cos(60° + 459)

Question 9.
cot 2A + tan A = cosec 2A.
Answer:
L.H.S. = cot2A + tanA

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:

Question 13.
cos(120° + A) + cos(120° – A) = -cos A
Answer:
L.H.S = cos(120° + A) + cos(120° – A)

Question 14.
\(\frac{\sin (A+B)}{\sin (A-B)} \quad \frac{\tan A+\tan B}{\tan A-\tan B}\)
Answer: