{"id":17988,"date":"2020-01-04T16:11:10","date_gmt":"2020-01-04T10:41:10","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17988"},"modified":"2021-07-02T15:31:13","modified_gmt":"2021-07-02T10:01:13","slug":"2nd-puc-maths-question-bank-chapter-2-ex-2-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-2-ex-2-2\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.2"},"content":{"rendered":"

Students can Download Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.2<\/h2>\n

2nd PUC Maths Inverse Trigonometric Functions NCERT Text Book Questions and Answers Ex 2.2<\/h3>\n

Question 1.
\n\\(3 \\sin ^{-1} x=\\sin ^{-1}\\left(3 x-4 x^{3}\\right), \\quad x \\in\\left[-\\frac{1}{2}, \\frac{1}{2}\\right]\\)
\nAnswer:
\nLet sin-1<\/sup> x be \u03b8 \u21d2 x = sin \u03b8
\nIn RHS putting the value of x
\nsin-1<\/sup> (3x – 4x3<\/sup>) \u21d2 sin-1<\/sup> (3 sin\u03b8 – 4 sin3<\/sup>\u03b8)
\n= sin-1<\/sup> (sin 3\u03b8)
\n[\u2235 3sin\u03b8 – 4sin3<\/sup> \u03b8 = sin3\u03b8]
\n= 3\u03b8
\nputting the value of \u03b8
\n3\u03b8 = 3 sin-1<\/sup> x = LHS
\n\u2234 3sin-1<\/sup> x = sin-1<\/sup> (3x – 4x3<\/sup>)
\nHence proved.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\n\\(3 \\cos ^{-1} x=\\cos ^{-1}\\left(4 x^{3}-3 x\\right), x \\in\\left[\\frac{1}{2}, 1\\right]\\)
\nAnswer:
\nLet cos-1<\/sup> x be \u03b8\u00a0\u21d2 x = cos \u03b8
\nIn RHS, putting the value of x
\ncos-1<\/sup> (4x3<\/sup> – 3x) \u21d2 cos1<\/sup> (4cos3<\/sup> \u03b8 – 3 cos\u03b8)
\n\u21d2 cos-1<\/sup> (cos 3\u03b8)
\n[\u2235 4cos3<\/sup> \u03b8 – 3cos\u03b8 = cos 3\u03b8]
\n= 30
\nputting the value of 0 = 30
\n\u21d2 3 cos-1<\/sup> x = LHS
\nLHS = RHS
\nHence proved<\/p>\n

Question 3.
\n\\(\\tan ^{-1} \\frac{2}{11}+\\tan ^{-1} \\frac{7}{24}=\\tan ^{-1} \\frac{1}{2}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\n\\(2 \\tan ^{-1} \\frac{1}{2}+\\tan ^{-1} \\frac{1}{7}=\\tan ^{-1} \\frac{31}{17}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Write the following function in the simplest form:<\/p>\n

Question 5.
\n\\(\\tan ^{-1} \\frac{\\sqrt{1+x^{2}}-1}{x}, x \\neq 0\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\n\\(\\tan ^{-1} \\frac{1}{\\sqrt{x^{2}-1}},|x|>1\\)
\nAnswer:
\nLet cosec-1<\/sup> x = 0
\n\u21d2 x = cosec 0
\n\"2nd<\/p>\n

Question 7.
\n\\(\\tan ^{-1}(\\sqrt{\\frac{1-\\cos x}{1+\\cos x}}), 0<x<\\pi\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 8.
\n\\(\\tan ^{-1}\\left(\\frac{\\cos x-\\sin x}{\\cos x+\\sin x}\\right), \\frac{-\\pi}{4}<x<\\frac{3 \\pi}{4}\\)
\nAnswer:
\nDivide each term of numerator and denominator inside the brackets by cos x
\n\\(\\Rightarrow \\tan ^{-1}\\left[\\frac{(\\cos x-\\sin x) \/ \\cos x}{(\\cos x+\\sin x) \/ \\cos x}\\right] \\Rightarrow \\tan ^{-1}\\left[\\frac{1-\\tan x}{1+\\tan x}\\right]\\)
\n\"2nd<\/p>\n

Question 9.
\n\\(\\tan ^{ -1 } \\frac { { x } }{ \\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } ,|{ x }|<{ a }\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\n\\(\\tan ^{-1}\\left(\\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\\right), a>0 ; \\frac{-a}{\\sqrt{3}} \\leq x \\leq \\frac{a}{\\sqrt{3}}\\)
\nAnswer:
\n\"2nd<\/p>\n

Find the values of each of the following<\/p>\n

Question 11.
\n\\(\\tan ^{-1}\\left[2 \\cos \\left(2 \\sin ^{-1} \\frac{1}{2}\\right)\\right]\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 12.
\ncot (tan-1<\/sup> a + cot-1<\/sup> a)
\nAnswer:
\n\"2nd<\/p>\n

Question 13.
\n\\(\\begin{aligned}&\\tan \\frac{1}{2}\\left[\\sin ^{-1} \\frac{2 x}{1+x^{2}}+\\cos ^{-1} \\frac{1-y^{2}}{1+y^{2}}\\right]\\\\&|x|<1, y>0 \\text { and } x y<1\\end{aligned}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nIn \\(\\sin \\left(\\sin ^{-1} \\frac{1}{5}+\\cos ^{-1} x\\right)=1\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 15.
\n\\(\\text { If } \\tan ^{-1} \\frac{x-1}{x-2}+\\tan ^{-1} \\frac{x+1}{x+2}=\\frac{\\pi}{4}\\)Then find the value of x
\nAnswer:
\n\"2nd
\n\"2nd
\nFind the values of each of the expressions in Exercises 16 to 18.<\/p>\n

Question 16.
\n\\(\\sin ^{-1}\\left(\\sin \\frac{2 \\pi}{3}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 17.
\n\\( \\tan ^{-1}\\left(\\tan \\frac{3 \\pi}{4}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 18.
\n\\(\\tan \\left(\\sin ^{-1} \\frac{3}{5}+\\cot ^{-1}\\left(\\frac{3}{2}\\right)\\right)\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Choose the correct Answer:<\/p>\n

Question 19.
\n\\(\\cos ^{-1}\\left(\\cos \\frac{7 \\pi}{6}\\right) \\text { is equal to }\\)
\n(A) \\(\\frac{7 \\pi}{6}\\)
\n(B) \\(\\frac{5 \\pi}{6}\\)
\n(C) \\(\\frac{\\pi}{3}\\)
\n(D) \\(\\frac{\\pi}{6}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 20.
\n\\(\\sin \\left(\\frac{\\pi}{3}-\\sin ^{-1}\\left(-\\frac{1}{2}\\right)\\right)\\)is equal to
\n(A) \u03c0
\n(B) \\(-\\frac{\\pi}{2}\\)
\n(C) 0
\n(D) \\(2 \\sqrt{3}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 21.
\n\\(\\tan ^{-1} \\sqrt{3}-\\cot ^{-1}(-\\sqrt{3})\\) is equal to
\n(A) \u03c0
\n(B) \\(-\\frac{\\pi}{2}\\)
\n(C) 0
\n(D) \\(2 \\sqrt{3}\\)
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. 2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.2 2nd PUC Maths Inverse …<\/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\/17988"}],"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=17988"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17988\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17988"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17988"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17988"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}