{"id":1599,"date":"2020-09-15T06:08:52","date_gmt":"2020-09-15T00:38:52","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1599"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-sslc-class-10-maths-solutions-chapter-11-ex-11-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-11-ex-11-2\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2<\/h2>\n

Question 1.
\nEvaluate the following :
\ni) sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0
\nii) 2 tan2<\/sup> 45\u00b0 + cos2 30\u00b0 – sin2<\/sup> 60\u00b0
\n\"KSEEB
\nSolution:
\ni) sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0
\n\"KSEEB<\/p>\n

ii) 2 tan2<\/sup> 45\u00b0 + cos2 30\u00b0 – sin2<\/sup>60\u00b0
\n= 2(tan 45\u00b0)2<\/sup> + (cos 30\u00b0)2<\/sup> – (sin 60\u00b0)2<\/sup>
\n= 2 (1)2<\/sup> + \\(\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}\\)
\n= 2 \u00d7 1
\n= 2
\n\"KSEEB
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 2.
\nChoose the correct option and justify your choice :
\ni) \\(\\frac{2 \\tan 30^{\\circ}}{1+\\tan ^{2} 30^{\\circ}}\\) =
\nA) sin 60\u00b0
\nB) cos 60\u00b0
\nC) tan 60\u00b0
\nD) sin 30\u00b0
\nSolution:
\nA) sin 60\u00b0
\n\"KSEEB<\/p>\n

ii) \\(\\frac{1-\\tan ^{2} 45^{\\circ}}{1+\\tan ^{2} 45^{\\circ}}\\) =
\nA) tan 90\u00b0
\nB) 1
\nC) sin 45\u00b0
\nD) 0
\nSolution:
\nD) 0
\n\"KSEEB<\/p>\n

iii) sin 2A = 2 sin A is true when A ;
\nA) 0\u00b0
\nB) 30\u00b0
\nC) 45\u00b0
\nD) 60\u00b0
\nSolution:
\nA) 0\u00b0
\nLHS = sin 2A = sin 0\u00b0 = 0
\nRHS = 2sin A = 2.sin0\u00b0 = 0<\/p>\n

iv) \\(\\frac{2 \\tan 30^{\\circ}}{1-\\tan ^{2} 30^{\\circ}}\\) =
\nA) cos 60\u00b0
\nB) sin 60\u00b0
\nC) tan 60\u00b0
\nD) sin 30\u00b0
\nSolution:
\nC) tan 60\u00b0
\n\"KSEEB
\n\"KSEEB<\/p>\n

Question 3.
\nIf tan (A + B) =\\(\\sqrt{3}\\) and tan (A – B) = \\(\\frac{1}{\\sqrt{3}}\\) 0\u00b0 < A + B \u2264 90\u00b0; A > B. find A and B.
\nSolution:
\ntan (A + B) = \\(\\sqrt{3}\\)
\ntan (A + B) = tan 60\u00b0
\nA + B = 60\u00b0
\ntan (A – B) = \\(\\frac{1}{\\sqrt{3}}\\) = tan 30\u00b0
\ntan(A – B) = tan 30\u00b0
\nA – B = 30\u00b0 \u2192 (2)
\nAdding (1) and (2)
\nA + B + A – B = 60 + 30
\n2A = 90
\nA = \\(\\frac{90}{2}\\) = 45\u00b0
\nA = 45\u00b0
\nPut A = 45\u00b0 in eqn (1)
\nA + B = 60
\nB = 60 – A= 60 – 45\u00b0
\nB = 15\u00b0.<\/p>\n

Question 4.
\nState whether the following are true or false. Justify your answer.
\ni) sin (A + B) = sin A + sin B
\nSolution:
\nFalse
\nTake A = 45\u00b0 and B = 45\u00b0
\nLHS: sin (45\u00b0 + 45\u00b0) = sin 90\u00b0 = 1
\nRHS: sin 45 + sin 45 = \\(\\frac{1}{\\sqrt{2}}+\\frac{1}{\\sqrt{2}}=\\frac{2}{\\sqrt{2}}\\)
\nLHS \u2260 RHS.<\/p>\n

ii) The value of sin \u03b8 increases as \u03b8 increases.
\nSolution:
\nTrue.
\n\"KSEEB
\n\u2234 The value of sin \u03b8 increases as \u03b8 increases.<\/p>\n

iii) The value of cos \u03b8 increases as \u03b8 increases.
\nSolution:
\nFalse.
\ncos 0\u00b0 = 1 cos 30\u00b0 = \\(\\frac{\\sqrt{3}}{2}\\) =0.87
\ncos 45\u00b0 = \\(\\frac{1}{\\sqrt{2}}\\) = 0.7 cos 60\u00b0 = \\(\\frac{1}{2}=\\) = 0.5
\ncos 90\u00b0 = 1
\n\u2234 The value of cos 0 increases as 0 increases – the statement is False.<\/p>\n

iv) sin \u03b8 = cos \u03b8 for all values of \u03b8.
\nSolution:
\nFalse.
\nsin 30\u00b0 = \\(\\frac{1}{2}\\) cos 30\u00b0 = \\(\\frac{\\sqrt{3}}{2}\\)
\nsin 30\u00b0 \u2260 cos 30\u00b0
\nBut sin 45\u00b0 = cos 45\u00b0 = \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

v) cot A is not defined for A = 0\u00b0.
\nSolution:
\nTrue.
\n\"KSEEB
\n\u2234 When A = 0\u00b0, cot A is not defined.<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2. Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.2 Question 1. Evaluate …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1599"}],"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\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1599"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1599\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1599"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1599"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}