{"id":20536,"date":"2020-01-21T17:45:18","date_gmt":"2020-01-21T12:15:18","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20536"},"modified":"2021-07-02T15:28:02","modified_gmt":"2021-07-02T09:58:02","slug":"2nd-puc-basic-maths-question-bank-chapter-14-ex-14-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-14-ex-14-3\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3"},"content":{"rendered":"

Students can Download Basic Maths Exercise 14.3 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 One Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nExpress each of the following a sum or difference of two trigonometric functions.
\n(i) sin 5A \u00b7 cos3A
\n(ii) cos 4A sin 2A
\n(iii) cos \\(\\left(\\frac{5 \\theta}{2}\\right)\\) . sin \\(\\left(\\frac{\\theta}{2}\\right)\\)
\n(iv) 2 cos 70\u00b0 cos 10\u00b0
\nAnswer:
\n(i) Using transformation formulae
\nsin5A . cos3A = \\(\\frac { 1 }{ 2 }\\)[sin(5A +3A) + sin(5A – 3A)]
\n= \\(\\frac { 1 }{ 2 }\\)(sin 8A + sin 2A)<\/p>\n

(ii) cos4A – sin2A = \\(\\frac { 1 }{ 2 }\\) [sin 6A – sin 2A]<\/p>\n

(iii) cos \\(\\left(\\frac{5 \\theta}{2}\\right)\\) . sin \\(\\left(\\frac{\\theta}{2}\\right)\\) = \\(\\frac { 1 }{ 2 }\\)[sin 3\u03b8 \u2013 sin 2\u03b8]<\/p>\n

(iv) 2cos 70\u00b0 . cos10\u00b0 = 2 x \\(\\frac { 1 }{ 2 }\\) [cos80\u00b0 + cos60\u00b0]
\n= cos80\u00b0 + \\(\\frac { 1 }{ 2 }\\)<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nExpress each of the following as the product of two trigonometric reactions.
\n(i) sin 12x + sin 4x
\n(ii) sin 7A – sin 3A
\n(iii) cos 2\u03b8 + cos 6\u03b8
\n(iv) sin 80\u00b0 – sin 40\u00b0
\nAnswer:
\n(i) sin12x + sin4x = 2.sin \\(\\frac{12 x+4 x}{2}\\) . cos \\(\\frac{12 x-4 x}{2}\\)<\/p>\n

(ii) sin7A \u2013 sin3A = 2 cos 5A.sin2A<\/p>\n

(iii) cos6\u03b8 + cos2\u03b8 = 2 cos4\u03b8.cos2\u03b8<\/p>\n

(iv) sin 80\u00b0 – sin 40\u00b0 = 2 cos60\u00b0. sin 20\u00b0 = 2.\\(\\frac { 1 }{ 2 }\\)sin 20\u00b0 = sin 20\u00b0<\/p>\n

Part – B<\/span><\/p>\n

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3Two Marks Questions and Answers<\/strong><\/p>\n

Prove the following<\/p>\n

Question 1.
\n\\(\\frac{\\cos 2 A-\\cos 12 A}{\\sin 12 A-\\sin 2 A}=\\tan 7 A\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\n\\(\\frac{\\sin x-\\sin y}{\\sin x \\sin y}=\\tan \\left(\\frac{x-y}{2}\\right) \\cdot \\cot \\left(\\frac{x+y}{2}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\n\\(\\frac{\\sin 2 \\alpha+\\sin 3 \\alpha}{\\cos 2 \\alpha-\\cos 3 \\alpha}=\\cot \\left(\\frac{\\alpha}{2}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\n\\(\\sin \\left(\\frac{\\pi}{3}+A\\right)-\\sin \\left(\\frac{\\pi}{3}-A\\right)=\\sin A\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 5.
\ncos A + cos (120\u00b0 – A) + cos(120\u00b0 + A) = 0
\nAnswer:
\nL.H.S. = cosA + cos A(120\u00b0 – A) + cos(120\u00b0 + A)
\n= cosA + 2cos 120\u00b0. cosA = cosA + 2cosA (-cos60\u00b0)
\n= cos A – 2 cos A . \\(\\frac { 1 }{ 2 }\\) = 0 = R.H.S.<\/p>\n

Question 6.
\nsin 65\u00b0 + cos 65\u00b0 = \\(\\sqrt{2} \\cos 20^{\\circ}\\)
\nAnswer:
\nL.H.S. sin65\u00b0 + cos65\u00b0
\n= sin(45\u00b0 + 20\u00b0) + cos(45\u00b0 + 20\u00b0)
\n\"2nd
\n= 2. \\(\\frac{1}{\\sqrt{2}}\\). cos20\u00b0= \\(\\sqrt{2}\\) cos20\u00b0 = R.H.S.<\/p>\n

Question 7.
\nIf A + B + C = \u03c0, Prove that tan A + tan B + tan C = tan A . tanB \u00b7 tan C
\nAnswer:
\nGiven A + B + C = \u03c0
\n\u21d2 A + B = \u03c0 – C
\ntan(A + B) = tan(180 – C)
\n\\(\\frac{\\tan A+\\tan B}{1-\\tan A \\cdot \\tan B}=-\\frac{\\tan C}{1}\\)
\ntanA + tanB = – tanc – tanB tanB \u00b7 tanc
\ntanA + tanB + tanC = tanA tanB tanC<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nProve that \\(\\frac{\\sin 2 A+\\sin 5 A-\\sin A}{\\cos 2 A+\\cos 5 A+\\cos A}=\\tan 2 A\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 9.
\nIf A + B + C = 180\u00b0. Prove that cot B. cot C + cot C \u00b7 cot A + cot A \u00b7 cot B = 1
\nAnswer:
\nGiven A + B + C = 180\u00b0; A + B = 180\u00b0 – C; cot(A + B) = cot (180\u00b0 – C)
\n\\(\\frac{\\cot (\\mathrm{A}) \\cdot \\cot (\\mathrm{B})-1}{\\cot \\mathrm{A}+\\cot \\mathrm{B}}=-\\cot \\mathrm{C}\\)
\ncotA . cotB – 1 = -cotA \u00b7 cotC – cot B . cotC.
\n\u2234 cotA . cotB + cotB cotC + cotC . cotA = 1.<\/p>\n

Question 10.
\nIf A + B + C = \\(\\frac{\\pi}{2}\\) . P.T tanA \u00b7 tanB + tan B . tan C + tan C \u00b7 tan A = 1
\nAnswer:
\nGiven A+B+C = \\(\\frac{\\pi}{2}\\); A + B = 90\u00b0 – C; tan(A + B) = tan(90\u00b0 – C)
\n\\(\\frac{\\tan A+\\tan B}{1-\\tan A \\tan B}=\\cot C=\\frac{1}{\\tan C}\\)
\n\u21d2 tanA tanC + tanB tanC = 1 – tanA \u00b7 tanB
\n\u21d2 tan A tanB + tanB tanC + tanC tanA = 1.<\/p>\n

Part – C<\/span><\/p>\n

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.3 Five Marks Questions and Answers<\/strong><\/p>\n

Prove the following<\/p>\n

Question 1
\n\\(\\frac{\\cos 7 x+\\cos 3 x-\\cos 5 x-\\cos x}{\\sin 7 x-\\sin 3 x-\\sin 5 x+\\sin x}=\\cot 2 x\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\ncos10\u00b0.cos30\u00b0. cos50\u00b0.cos 70\u00b0 = \\(\\frac { 3 }{ 16 }\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\ncos 20\u00b0 \u00b7 cos40\u00b0 \u00b7 cos60\u00b0 . cos80\u00b0 = \\(\\frac { 1 }{ 16 }\\)
\nAnswer:
\nL.H.S. = cos 20\u00b0.cos40\u00b0.cos60\u00b0.cos80\u00b0
\n= cos60\u00b0.cos80\u00b0.cos40\u00b0.c0s60\u00b0
\n\"2nd<\/p>\n

Question 4.
\nIf A + B + C= 180\u00b0C. Prove that sin2<\/sup>a + sin2<\/sup> \\(\\left(\\frac{2 \\pi}{3}+A\\right)\\) + sin \\(\\left(\\frac{2 \\pi}{3}-A\\right)\\) = \\(\\frac { 3 }{ 2 }\\)
\nAnswer:
\nGiven A + B + C = 180\u00b0 \\(\\left[\\because \\frac{2 \\pi}{3}=120^{\\circ}\\right]\\)
\nsin2<\/sup>A + sin2<\/sup>(120\u00b0 + A) + sin2<\/sup>(120\u00b0 – A)
\n\"2nd<\/p>\n

Question 5.
\nsin 2A + sin 2B + sin 2C = 4 sin A \u00b7 sinB \u00b7 sin C
\nAnswer:
\nL.H.S. = sin2A + sin2B + sin2C
\n= 2sin(A + B) \u00b7 cos(A – B) + 2sinC cosC
\n= 2sinC.cos(A – B) + 2 sinC cosC = 2sinC [cos(A – B) – cos(A + B)]
\n= 2 sinC[-2 sinA \u00b7 sin(-B)]
\n= 4sinA . sinB . sinC \u2235 sin(-B)
\n= R.H.S. = SinB.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\ntan 2A + tan 2B + tan 2C = tan 2A . tan 2B . tan 2C
\nAnswer:
\nGiven
\nA + B + C = 180\u00b0
\n\u21d2 2A + 2B + 2C = 360\u00b0
\n2A + 2B = 360\u00b0 – 2C
\ntan(2A + 2B) = tan(360\u00b0 – 2C)
\n\\(\\frac{\\tan 2 A+\\tan 2 B}{1-\\tan 2 A \\cdot \\tan 2 B}=\\frac{-\\tan 2 C}{1}\\)
\ntan2A + tan2B = – tan2C + tan2A tan2B tan2C
\ntan2A + tan2B + tan2C = tan 2A tan2B \u00b7 tan2C<\/p>\n

Question 7.
\nsin 4A + sin 4B + sin 4C = -4sin2A . sin2B \u00b7 sin2C
\nAnswer:
\nL.H.S = sin4A + sin4B + sin4C [A+B+C = 180]
\n= 2 sin(2A + 2B) \u00b7 cos(2A – 2B) + 2 sinc.cos2C [2A + 2B + 2C = 360]
\n= (-sin2C) \u00b7 cos(2A – 2B) + (2 sin2C \u00b7 cos2C) [2A + 2B = 360 – 20]
\n= -2 sin 2C [cos(2A – 2B) \u2013 cos2C] [sin(2A + 2B) = – sin2C]
\n= – 2 sin 2C [cos(2A – 2B) – cos(2A + 2B)] [cos(2A + 2B) = cos 2C]
\n= -2 sin2C[-2 sin2A \u00b7 sin(-2B)]
\n= – 4 sin2A \u00b7 sin2B \u00b7 sin2C
\n= R.H.S.<\/p>\n

Question 8.
\ncos2<\/sup>A + cos2<\/sup>B – cos2<\/sup>C = 1 – 2 sin A . sin B \u00b7 sin C
\nAnswer:
\nL.H.S. = cos2<\/sup>A + cos2<\/sup>B – cos2<\/sup>C
\n\"2nd
\n= \\(\\frac { 1 }{ 2 }\\)[1+2cos(A+B) cos(A – B) – (2cos2<\/sup>C – 1)]
\n= \\(\\frac { 1 }{ 2 }\\)[1+2cos(A+B).cos(A – B) \u2013 2cos2<\/sup>C +1]
\n= \\(\\frac { 1 }{ 2 }\\)[2 + 2(-cosc) cos(A – B) 2cos2<\/sup>C]
\n= \\(\\frac { 1 }{ 2 }\\)[2 – 2 cos C[cos (A – B) + cos C]
\n= \\(\\frac { 1 }{ 2 }\\)[2 – 2cosC[cos(A – B) \u2013 cos(A+B)]]
\n= 1 – cosC[-2sin A . sin(-B)]
\n=1 – cosC[2sin A sin B]
\n= 1 – 2sinA sinB cosC = R.H.S.<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\n\\(\\frac{\\sin 2 A+\\sin 2 B+\\sin 2 C}{\\sin 2 A+\\sin 2 B-\\sin 2 C}=\\tan A \\cdot \\tan B\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 10.
\ncos2A + cos2B + cos2C = -1 – 4 cosA cosB cosC
\nAnswer:
\nL.H.S. = cos2A + cos2B + cos2C
\n= 2cos(A + B) \u00b7 cos(A – B) + 2 cos2<\/sup>C \u2013 1
\n= -2cosC . cos(A – B) + 2 cos2<\/sup>C \u2013 1
\n= -2 cosC[cos (A – B) – cos C] – 1
\n= -2 cosC{cos (A – B) + cos(A + B)} – 1
\n= -2 cosC (2 cosA \u00b7 cosB) – 1
\n= – 1 – 4 cos A cosB cosC = R.H.S.<\/p>\n

Question 11.
\nsin2<\/sup>A + sin2<\/sup>B – sin2<\/sup>C = 2sinA \u00b7 sin B. cosC
\nAnswer:
\nL.H.S. = sin2<\/sup>A + sin2<\/sup>B – sin2<\/sup>C
\n\"2nd
\n= \\(\\frac { 1 }{ 2 }\\)[1-(cos 2A + cos2B \u2013cos2C)]
\n= \\(\\frac { 1 }{ 2 }\\)[1-{{2cos(A+B). Cos(A – B) \u2013 2 cos2<\/sup>C+1}]
\n\"2nd
\n= \\(\\frac { 1 }{ 2 }\\)[+2cosC . cos(A – B) +2cos2<\/sup>c]
\n= \\(\\frac { 1 }{ 2 }\\)[2cosC(cos(A – B)+cosC]
\n= \\(\\frac { 1 }{ 2 }\\)[2cosC(cos(A – B) – cos(A + B))]
\n= \\(\\frac { 1 }{ 2 }\\)[2cosC[-2sin A.sin(-B)]]
\n= 2 sinA sinB sinc<\/p>\n

Question 12.
\nIf A + B + C = \u03c0. Prove that:
\ncos 2A + cos 2B – cos2C = 1 – 4 sin A . sin B \u00b7 cos C
\nAnswer:
\nL.H.S. = cos2A + cos2B – cos2C
\n= 2cos(A + B) cos(A – B) – (2 cos2<\/sup>C – 1)
\n= -2cosC Cos(A – B) \u2013 2 cos2<\/sup>C + 1 [\u2235 cosC=-cos(A+B)]
\n= 1 – 2cos C[cos(A – B) – cos (A + B)]
\n= 1 – 2cos C[-2 sinA . sin(- B)]
\n= 1 – 4 sin A sin B cos C = R.H.S.<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nIf A + B + C = 180\u00b0. Prove that:
\nsin 2A – sin 2B + sin 2C = 4 cos A \u00b7 sin B \u00b7 cos C
\nAnswer:
\nGiven A + B + C = 180\u00b0
\n2A + 2B + 2C = 360\u00b0
\n2A + 2B = 360\u00b0 – 2C
\nsin(2A + 2B) = sin(360\u00b0 – 20) = – sin2C
\ncos(2A + 2B) = cos(360\u00b0 – 2C) = cos 2C
\nL.H.S = sin2A – sin2B + sin2C
\n= 2cos(A + B) \u00b7 sin(A – B) + 2sinC. cosC
\n= – 2cosC.sin(A – B) + 2 sinc.cosC
\n= 2 cosC [sinC – sin (A – B)]
\n= 2 cosC [sin(A + B) – sin(A – B)]
\n= 2 cos C [2cosA . sinB]
\n= 4 cosA sinB.cosC = R.H.S.<\/p>\n

Question 14.
\ncos 2A – cos 2B + cos 2C = 1 – 4 sin A \u00b7 cos B \u00b7 sin C
\nAnswer:
\nL.H.S. = cos 2A – cos2B + cos2C
\n= -2 sin(A + B) . sin(A – B) + 1 – 2sin2<\/sup>C
\n= 1 – 2 sinC sin (A – B) – 2 sin2<\/sup>C [\u2235 sin(A+B)-sinc]
\n= 1 – 2 sinC [sin(A – B) + sinC]
\n= 1 – 2 sinC [sin(A + B) + sin(A – B)]
\n= 1 – 2 sinC [2sinA . cosB]
\n= 1 – 4 sinA cosB sinC = R.H.S.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 14.3 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 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20536"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=20536"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20536\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20536"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20536"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20536"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}