{"id":20484,"date":"2020-01-21T15:32:27","date_gmt":"2020-01-21T10:02:27","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20484"},"modified":"2021-07-02T15:28:08","modified_gmt":"2021-07-02T09:58:08","slug":"2nd-puc-basic-maths-question-bank-chapter-14-ex-14-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-14-ex-14-2\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 14 Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.2"},"content":{"rendered":"

Students can Download Basic Maths Exercise 14.2 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 14.2<\/h2>\n

Part – A & B<\/span><\/p>\n

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.2 One or Two Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nIf sin A = \\(\\frac { 1 }{ 2 }\\) find sin 2A
\nAnswer:
\nGiven sin A = \\(\\frac { 1 }{ 2 }\\) \u21d2 cos A = \\(\\frac{\\sqrt{3}}{2}\\) sin2A
\n\"2nd<\/p>\n

Question 2.
\nIf cos A = \\(\\frac{\\sqrt{3}}{2}\\) find cos 2A
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

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

Question 4.
\nIf sin A = \\(\\frac { 3 }{ 5 }\\) find sin 3A
\nAnswer:
\nGiven sin A = \\(\\frac { 3 }{ 5 }\\) sin3A = 3sinA – 4sin3<\/sup>A = \\(3 \\cdot \\frac{3}{5}-4\\left(\\frac{3}{5}\\right)^{3}\\)
\n\\(=\\frac{9}{5}-4 \\cdot \\frac{27}{125}=\\frac{225-108}{125}=\\frac{117}{125}\\)<\/p>\n

Question 5.
\nIf cos A = \\(\\frac { 4 }{ 5 }\\) 4 find cos 3A
\nAnswer:
\nGiven cos A = \\(\\frac { 4 }{ 5 }\\)
\n\"2nd<\/p>\n

Question 6.
\nIf tan A = \\(\\frac { 3 }{ 4 }\\) find tan 3A
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nFind the value of 3 sin 10\u00b0 – 4 sin3<\/sup>10\u00b0
\nAnswer:
\n3sin10\u00b0 – 4sin3<\/sup>10\u00b0 is of the form 3 sin A – 4sin3<\/sup>A = sin3A = sin 3-10\u00b0 = sin 30\u00b0 = \\(\\frac { 1 }{ 2 }\\)<\/p>\n

Question 8.
\nIf cot A = \\(\\frac { 12 }{ 5 }\\) and A is acute find sin3A
\nAnswer:
\nGiven cot A = \\(\\frac { 12 }{ 5 }\\) \u21d2 sin A = \\(\\frac { 5 }{ 13 }\\) cosA = \\(\\frac { 12 }{ 13 }\\)
\n\u2234 sin 3A = 3sinA – 4 sin3<\/sup>A
\n\"2nd<\/p>\n

Question 9.
\nShow that tan A = \\(\\frac{\\tan (A-B)+\\tan B}{1-\\tan (A-B) \\tan B}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 10.
\nProve that \\(\\frac{\\cos 2 A}{1+\\sin 2 A}=\\frac{\\cos A-\\sin A}{\\cos A+\\sin A}\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nProve that \\(\\frac{1+\\sin 2 \\theta}{\\cos 2 \\theta}=\\frac{1+\\tan \\theta}{1-\\tan \\theta}\\)
\nAnswer:
\n\"2nd<\/p>\n

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

Question 13.
\nProve that \\(\\frac{\\sin 2 \\theta}{1+\\cos 2 \\theta}=\\tan \\theta\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 14.
\nProve that (sin A – cos A)2<\/sup> = 1 – sin 2A
\nAnswer:
\nL.H.S (sinA – cosA)2<\/sup>
\n= sin2<\/sup>A + cos2<\/sup>A \u2013 2sinA cosA
\n= 1 – sin2A = R.H.S.<\/p>\n

\"KSEEB<\/p>\n

Question 15.
\nProve that cos4<\/sup>\u03b8 – sin4<\/sup>\u03b8 = 2 cos2<\/sup>\u03b8 – 1
\nAnswer:
\nL.H.S. cos4<\/sup>\u03b8 – sin4<\/sup>\u03b8
\n= (cos2<\/sup>\u03b8)2<\/sup> – (sin2<\/sup>\u03b8)2<\/sup>
\n= (cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8) (cos2<\/sup>\u03b8 – sin2<\/sup>\u03b8)
\n= 1 . (cos2<\/sup>\u03b8 – (1 – cos2<\/sup>\u03b8)
\n= cos2<\/sup>\u03b8 – 1 + cos2<\/sup>\u03b8 = 2cos2<\/sup>\u03b8 – 1 = R.H.S.<\/p>\n

Question 16.
\nProve that \\(\\frac{\\cos ^{3} A-\\sin ^{3} A}{\\cos A-\\sin A}=1+\\frac{1}{2} \\sin 2 A\\)
\nAnswer:
\n\"2nd<\/p>\n

Part – C<\/span><\/p>\n

2nd PUC Basic Maths Compound Angles, Multiple Angles, Submultiple Angles & Transformation Formulae Ex 14.2Five Marks Question and Answers<\/strong><\/p>\n

Prove the following<\/p>\n

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

Question 2.
\n\\(\\frac{\\cos 3 A}{2 \\cos 2 A-1}=\\cos A\\) and hence find cos 15\u00b0
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\n\\(\\frac{1-\\cos 2 A+\\sin 2 A}{1+\\cos 2 A+\\sin 2 A}=\\tan A\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\ncos6<\/sup>A+ sin6<\/sup>A = 1 – \\(\\frac { 3 }{ 4 }\\) sin3<\/sup>(2A)
\nAnswer:
\ncos6<\/sup>A+ sin6<\/sup>A [\u2235 a3<\/sup> + b3<\/sup> = (a + b)3<\/sup> \u2013 3ab(a+b)]
\n= (cos2<\/sup>A)3<\/sup> + (sin2<\/sup>A)3<\/sup>
\n= (cos2<\/sup>A + sin2<\/sup>A)3<\/sup> – 3cos2<\/sup>A – sin2<\/sup>A(cos3<\/sup>A + sin3<\/sup>A)
\n= 13<\/sup> – 3cos2<\/sup>A \u00b7 sin2<\/sup>A.1
\n\"2nd<\/p>\n

Question 5.
\nProvfe that \\(\\frac{\\sin 3 \\theta}{\\sin \\theta}-\\frac{\\cos 3 \\theta}{\\cos \\theta}=2\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\nsec (45\u00b0 + A) \u00b7 sec(45\u00b0 – A) = 2 sec 2A
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

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

Question 8.
\nProve that \\(\\frac{\\cos 2 A}{1+\\sin 2 A}\\) = tan(45\u00b0 – A)
\nAnswer:
\n\"2nd<\/p>\n

Question 9.
\nIf tan \u03b1 = \\(\\frac { 1 }{ 3 }\\), tan \u03b2 = \\(\\frac { 1 }{ 7 }\\) P.T. tan(2\u03b1 + \u03b2) = 45\u00b0.
\nAnswer:
\nL.H.S = tan(2\u03b1 + \u03b2)
\n\"2nd
\ntan(2\u03b1 + \u03b2) = 1 = tan 45\u00b0 = R.H.S.<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nIf tan2<\/sup>(45\u00b0 + \u03b8) = \\(\\frac { a }{ b }\\) prove that \\(\\frac{b-a}{b+a}\\) = – sin 2\u03b8
\nAnswer:
\n\"2nd<\/p>\n

Question 11.
\nProve that tan 2\u03b8 – tan\u03b8 = tan \u03b8 . sec 2\u03b8.
\nAnswer:
\nL.H.S = tan2\u03b8 – tan\u03b8
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nProve that cos 2\u03b1 – tan \u03b1 = \\(\\frac{\\cos 3 \\alpha}{\\cos \\alpha \\cdot \\sin 2 \\alpha}\\)
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Exercise 14.2 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\/20484"}],"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=20484"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20484\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20484"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20484"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20484"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}