{"id":20443,"date":"2020-01-21T12:45:05","date_gmt":"2020-01-21T07:15:05","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20443"},"modified":"2021-07-02T15:28:11","modified_gmt":"2021-07-02T09:58:11","slug":"2nd-puc-basic-maths-question-bank-chapter-14-ex-14-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-14-ex-14-1\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1"},"content":{"rendered":"

Students can Download Basic Maths Exercise 14.1 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 14.1<\/h2>\n

Part – A<\/span><\/p>\n

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 Two Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nObtain the values of the trigonometric functions of 75\u00b0, 15\u00b0 and 105\u00b0 and prove that
\n(i) tan 75\u00b0 + cot 75\u00b0 = 4
\n(ii) sin 105\u00b0 + cos 105\u00b0 = \\(\\frac{1}{\\sqrt{2}}\\)
\n(iii) Sec 15\u00b0 + cosec 15\u00b0 = \\(2 \\sqrt{6}\\)
\nAnswer:
\nsin(45\u00b0 + 30\u00b0) = sin45\u00b0. cos30\u00b0 + cos45\u00b0 \u00b7 sin 30\u00b0 = \\(\\frac{1}{\\sqrt{2}} \\frac{\\sqrt{3}}{2}+\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{2}\\)
\n\u2234 sin 75\u00b0 = \\(\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\\)
\nsin15\u00b0 = sin(45\u00b0 – 30\u00b0) = \\(\\frac{\\sqrt{3}-1}{2 \\sqrt{2}}\\)
\nsin(105\u00b0) = sin(60\u00b0 + 45\u00b0) = sin60\u00b0. sin45\u00b0 + cos60\u00b0.cos45\u00b0
\n= \\(\\frac{\\sqrt{3}}{2} \\cdot \\frac{1}{\\sqrt{2}}+\\frac{1}{2} \\frac{1}{\\sqrt{2}}\\) ; sin 105\u00b0 = \\(\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\\)
\ncos(75\u00b0) = cos(45\u00b0 + 30\u00b0) = cos 45\u00b0 – cos30\u00b0 + sin45\u00b0 . sin30\u00b0 = \\(\\frac{\\sqrt{3}+1}{2 \\sqrt{2}}\\)
\ncos105\u00b0 = cos(60\u00b0 + 45\u00b0) = cos60\u00b0. cos45\u00b0 – sin60\u00b0 . sin45\u00b0 = \\(\\frac{+1-\\sqrt{3}}{2 \\sqrt{2}}\\)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nIf 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.
\nAnswer:
\nGiven sin A = \\(\\frac { 3 }{ 5 }\\) , cos B = \\(\\frac { 4 }{ 5 }\\)
\n\u21d2 cos A = \\(\\frac { 4 }{ 5 }\\) & sin B = \\(\\frac { 3 }{ 5 }\\) \u2235 sin2<\/sup>A + cos2<\/sup>A = 1
\n(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}\\)
\n(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\\)<\/p>\n

Question 3.
\nIf cos A = \\(\\frac { 5 }{ 13 }\\), cos B = \\(\\frac { 24 }{ 25 }\\) find cos(A + B) and sin (A – B). A and B are acute angles.
\nAnswer:
\nGiven
\ncos A = \\(\\frac { 5 }{ 13 }\\) cosB = \\(\\frac { 24 }{ 25 }\\)
\n\"2nd
\ncos (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}\\)
\nsin(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}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nIf sec A = \\(\\frac { 17 }{ 8 }\\) , cosec B = \\(\\frac { 5 }{ 4 }\\) find set (A + B). cosec(A + B), A and B are acute angles
\nAnswer:
\nGiven sec = \\(\\frac { 17 }{ 8 }\\) \u21d2 cos A = \\(\\frac { 8 }{ 17 }\\) \u2234 sin A = \\(\\frac { 15 }{ 17 }\\) cosec B = \\(\\frac { 5 }{ 4 }\\), sinB = \\(\\frac { 4 }{ 5 }\\), cosB = \\(\\frac { 3 }{ 5 }\\)
\n\"2nd
\nsin(A – B) = sinA cosB – cosA sinB
\n\\(=\\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 }\\)<\/p>\n

Question 5.
\nIf sin A = \\(\\frac { 3 }{ 5 }\\) cos B = \\(\\frac { -8 }{ 17 }\\) < A < \u03c0 and \\(\\frac{\\pi}{2}\\) < B < \u03c0. Find the values of sin(A + B) and cos(A – B).
\nAnswer:
\nGiven sin A = \\(\\frac { 3 }{ 5 }\\) . cos A = \\(\\frac { -4 }{ 5 }\\)
\ncos B = \\(\\frac { -8 }{ 17 }\\), sin B = \\(\\frac { 15 }{ 17 }\\)
\nsin(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}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIf sin A = \\(\\frac { 7 }{ 25 }\\), cos B = \\(\\frac { -12 }{ 13 }\\) where \\(\\frac{\\pi}{2}\\) < A < \u03c0 and \u03c0 < B <\u00a0 \\(\\frac{3 \\pi}{2}\\) find the values of:
\n(i) sin(A + B)
\n(ii) cos(A + B)
\n(iii) sin(A – B)
\n(ii) cos(A – B)
\n(v) tan(A + B)
\n(vi) tan(A – B)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nIf tan A = \\(\\frac { 1 }{ 2 }\\), tan B = \\(\\frac { 1 }{ 3 }\\) Find tan(A + B), tan(A – B)
\nAnswer:
\n\"2nd<\/p>\n

Question 8.
\ntan(A – B) = \\(\\frac { 1 }{ 7 }\\), tan A = \\(\\frac { 1 }{ 2 }\\) show that A + B = 45\u00b0.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nIf tan A = \\(\\frac { 1 }{ 3 }\\) tan (A + B) = \\(\\frac { 1 }{ 7 }\\) find tan B.
\nAnswer:
\n\"2nd<\/p>\n

Question 10.
\ntan A = \\(\\frac { 3 }{ 4 }\\) and tan B = \\(\\frac { 1 }{ 7 }\\) . show that tan (A + B) = 1.
\nAnswer:
\nGiven tan A = \\(\\frac { 3 }{ 4 }\\), tanB = \\(\\frac { 1 }{ 7 }\\)
\n\"2nd<\/p>\n

Question 11.
\nIf tan A = \\(\\frac { 5 }{ 6 }\\) and tan(A + B) = 1 Show that tanB = \\(\\frac { 1 }{ 11 }\\)
\nAnswer:
\nGiven
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nIf tan \u03b1 = \\(\\frac{\\mathrm{n}}{\\mathrm{n}+1}\\) and tan \u03b2 = \\(\\frac{1}{2 n+1}\\) Show that \u03b1 + \u03b2 = \\(\\frac{\\pi}{4}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 13.
\nProve that \\(\\frac{\\cos 2 A}{\\sec A}+\\frac{\\sin 2 A}{\\csc A}=\\cos A\\)
\nAnswer:
\nL.H.S. = cos2A . \\(\\frac{1}{\\sec A}\\) + sin 2A . \\(\\frac{1}{\\csc A}\\)
\n= cos2A . cosA + sin2A . sinA
\n= cos(2A – A) = cosA = R.H.S.<\/p>\n

Question 14.
\nProve that sin(45\u00b0 + A) + cos(45\u00b0 + A) = \\(\\sqrt{2}\\) cosA ,
\nAnswer:
\nL.H.S. = sin(45\u00b0 + A) + cos(45\u00b0 + A)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 15.
\nProve that cos \\(\\left(A+\\frac{\\pi}{4}\\right)=\\frac{1}{\\sqrt{2}}\\) (cos A – sin A).
\nAnswer:
\n\"2nd<\/p>\n

Question 16.
\nProve 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 }\\).
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Part – B<\/span><\/p>\n

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.1 Five Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nProve the following \\(\\frac{\\sin (A+B)}{\\cos A \\cdot \\cos B}\\) =tan A + tanB.
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\n\\(\\frac{\\sin (A-B)}{\\sin A \\sin B}=\\cot B-\\cot A\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\nShow that cos(A + B) cos(A – B) = cos2<\/sup>A – sin2<\/sup>B = cos2<\/sup>B – sin2<\/sup>A.
\nAnswer:
\nL.H.S = cos(A + B).cos(A – B)
\n= (cos A . cosB – sin A sinB) (cos A cosB + sinA sinB)
\n= cos2A . cos2B – sin2A \u00b7 sin 2B = cos2A(1 – sin2B) – (1 – cos2A) \u00b7 sin2B
\n\"2nd
\nSimilarly cos2A . cos2B – sin2A \u00b7 sin2B = (1 – sin2A) cos2B – sin2A(1 – cos2B)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nsin(A + B) sin(A – B) = cos2<\/sup>B – Cos2<\/sup>A
\nAnswer:
\nL.H.S = sin(A + B) sin(A – B) = (sinA cosB + cosA sinB) (sin A cosB – CosA sinB)
\n= sin2<\/sup>A cos2<\/sup>B – cos2<\/sup>A sin2<\/sup>B = (1 – Cos2<\/sup>A) cos2<\/sup>B – cos2<\/sup>A (1 – cos2<\/sup>B)
\n\"2nd<\/p>\n

Question 5.
\n\\(\\cos \\left(\\frac{\\pi}{4}-A\\right)-\\sin \\left(\\frac{\\pi}{4}+A\\right)=0\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\n\\(\\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\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\ntan 15\u00b0 + cot 15\u00b0 = 4.
\nAnswer:
\n\"2nd<\/p>\n

Question 8.
\nsin 105\u00b0 + cos 105\u00b0 = cos 45\u00b0
\nAnswer:
\nL.H.S = sin105\u00b0 + cos(105\u00b0) = sin(60\u00b0 + 45\u00b0) + cos(60\u00b0 + 459)
\n\"2nd<\/p>\n

Question 9.
\ncot 2A + tan A = cosec 2A.
\nAnswer:
\nL.H.S. = cot2A + tanA
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nP.T tan A tan 3A, tan 4A = tan 4A – tan3A – tanA.
\nAnswer:
\nConsider tan4A = tan(3A + A)
\n\\(\\frac{\\tan 4 A}{1}=\\frac{\\tan 3 A+\\tan A}{1-\\tan 3 A \\cdot \\tan A}\\) ; tan4A(1 – tan3A \u00b7 tanA) = tan3A + tanA
\ntan4A – tan4A tan3A tanA = tan3A + tanA
\ntan4A – tan3A – tanA = tan4A \u00b7 tan3A tanA<\/p>\n

Question 11.
\nProve that cos(45\u00b0 – A) \u00b7 cos(45\u00b0 – B) – sin(45\u00b0 – A) sin(45\u00b0 – B) = sin(A + B).
\nAnswer:
\nL.H.S = cos(45\u00b0 – A) \u00b7 cos(45\u00b0 – B) – sin(45\u00b0 – A) \u00b7 sin(45\u00b0 – B)
\n= cos(45\u00b0 \u2013 A + 45\u00b0 – B) \u2235 cosA . cosB – sinA . sinB
\n= cos(90\u00b0 – (A + B)) = cos (A + B) = sin(A + B) = R.H.S.<\/p>\n

Question 12.
\nShow that \\(\\sum \\frac{\\sin (A-B)}{\\cos A \\cdot \\cos B}=0\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 13.
\ncos(120\u00b0 + A) + cos(120\u00b0 – A) = -cos A
\nAnswer:
\nL.H.S = cos(120\u00b0 + A) + cos(120\u00b0 – A)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\n\\(\\frac{\\sin (A+B)}{\\sin (A-B)} \\quad \\frac{\\tan A+\\tan B}{\\tan A-\\tan B}\\)
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 14.1 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\/20443"}],"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=20443"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20443\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20443"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20443"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20443"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}