{"id":18086,"date":"2020-01-06T10:55:12","date_gmt":"2020-01-06T05:25:12","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=18086"},"modified":"2021-07-02T15:31:10","modified_gmt":"2021-07-02T10:01:10","slug":"2nd-puc-maths-question-bank-chapter-2-miscellaneous-ex","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-2-miscellaneous-ex\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise"},"content":{"rendered":"

Students can Download Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 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

Karnataka 2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise<\/h2>\n

Find the value of the following:<\/p>\n

Question 1.
\n\\(\\cos ^{-1}\\left(\\cos \\frac{13 \\pi}{6}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\n\\(\\tan ^{-1}\\left(\\tan \\frac{7 \\pi}{6}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Prove that<\/p>\n

Question 3.
\n\\(2 \\sin ^{-1} \\frac{3}{5}=\\tan ^{-1} \\frac{24}{7}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 4.
\n\\(\\sin ^{-1} \\frac{8}{17}+\\sin ^{-1} \\frac{3}{5}=\\sin ^{-1} \\frac{77}{85}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 5.
\n\\(\\cos ^{-1} \\frac{4}{5}+\\cos ^{-1} \\frac{12}{13}=\\cos ^{-1} \\frac{33}{65}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\n\\(\\cos ^{-1} \\frac{12}{13}+\\sin ^{-1} \\frac{3}{5}=\\sin ^{-1} \\frac{56}{65}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\n\\(\\tan ^{-1} \\frac{63}{16}=\\sin ^{-1} \\frac{5}{13}+\\cos ^{-1} \\frac{3}{5}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 8.
\n\\(\\tan ^{-1} \\frac{1}{5}+\\tan ^{-1} \\frac{1}{7}+\\tan ^{-1} \\frac{1}{3}+\\tan ^{-1} \\frac{1}{8}=\\frac{\\pi}{4}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Prove that<\/p>\n

Question 9.
\n\\(\\tan ^{-1} \\sqrt{x}=\\frac{1}{2} \\cos ^{-1}\\left(\\frac{1-x}{1+x}\\right), x \\in[0,1]\\)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\n\\(\\cot ^{-1}\\left(\\frac{\\sqrt{1+\\sin x}+\\sqrt{1-\\sin x}}{\\sqrt{1+\\sin x}-\\sqrt{1-\\sin x}}\\right)=\\frac{x}{2}, x \\in\\left(0, \\frac{\\pi}{4}\\right)\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 11.
\n\\(\\begin{aligned}&\\tan ^{-1}\\left(\\frac{\\sqrt{1+x}-\\sqrt{1-x}}{\\sqrt{1+x}+\\sqrt{1-x}}\\right)=\\frac{x}{4}-\\frac{1}{2} \\cos ^{-1} x\\\\&-\\frac{1}{\\sqrt{2}} \\leq x \\leq[\\text { Hint: put } x=\\cos 2 \\theta]\\end{aligned}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 12.
\n\\(\\frac{9 \\pi}{8}-\\frac{9}{4} \\sin ^{-1} \\frac{1}{3}=\\frac{9}{4} \\sin ^{-1} \\frac{2 \\sqrt{2}}{3}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 13.
\n2 tan-1<\/sup> (cos x) = tan-1<\/sup> (2cosec x)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\n\\(\\tan ^{-1} \\frac{1-x}{1+x}=\\frac{1}{2} \\tan ^{-1} x,(x>0)\\)
\nAnswer:
\n\"2nd
\n\"2nd
\nChoose the correct Answer<\/p>\n

Question 15.
\n\\(\\sin \\left(\\tan ^{-1} x\\right),|x|<1 \\text { is equal to }\\)
\n(A) \\(1, \\frac{1}{2}\\)
\n[B] \\(1, \\frac{1}{2}\\)
\n(C) 0
\n(D)\\(\\frac{1}{2}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 16.
\n\\(\\sin ^{-1}(1-x)-2 \\sin ^{-1} x=\\cfrac{\\pi}{2}, \\text { then } x\\)
\n(A) \\(0, \\frac{1}{2}\\)
\n[B] \\(1, \\frac{1}{2}\\)
\n(C) 0
\n(D)\\(\\frac{1}{2}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 17.
\n\\(\\tan ^{-1}\\left(\\cfrac{x}{y}\\right)-\\tan ^{-1} \\cfrac{x-y}{x+y} \\text { is equal to }\\)
\n(A)\\(\\frac{\\pi}{2}\\)
\n(B)\\(\\frac{\\pi}{3}\\)
\n(C)\\(\\frac{\\pi}{4}\\)
\n(D)\\(\\frac{-3 \\pi}{4}\\)
\nAnswer:
\n\"2nd<\/p>\n

2nd PUC Maths Inverse Trigonometric Functions Miscellaneous Exercise Additional Questions and Answers<\/h3>\n

Multiple Choice Questions:<\/strong><\/p>\n

Question 1.
\ncos-1<\/sup> cos(x) = x, then x lies in the interval
\n(a) [0,\u03c0]
\n(b) \\(\\left[\\frac{-\\pi}{2}, \\frac{\\pi}{2}\\right]\\)
\n(c) (o,\u03c0])
\n(d) \\(\\left(\\frac{-\\pi}{2}, \\frac{\\pi}{2}\\right)\\)
\nAnswer:
\n(a) [0,\u03c0]<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\n\\(\\cos ^{-1} \\cos \\left(\\cfrac{4 \\pi}{3}\\right)=\\)
\n(a) \\(\\frac{4 \\pi}{3}\\)
\n(b) \\(\\frac{2 \\pi}{3}\\)
\n(c) \\(\\frac{-\\pi}{3}\\)
\n(d) – \u03c0
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\n\\(\\sin ^{-1}\\left(\\cfrac{4}{5}\\right)-\\sec ^{-1}\\left(\\cfrac{5}{4}\\right)=\\sin ^{-1}\\left(\\cfrac{4}{5}\\right)+\\cos ^{-1}\\left(\\cfrac{4}{5}\\right)\\)
\n(a) \\(\\frac{\\pi}{2}\\)
\n(b) 0
\n(c) \u03c0
\n(d) \\(\\frac{-\\pi}{4}\\)
\nAnswer:
\n(a) sin1<\/sup> x + cos-1<\/sup> x = \\(\\frac{\\pi}{2}\\)<\/p>\n

Question 4.
\n\\(\\left(\\sqrt[12]{(2 x-2)}\\left(\\cfrac{-1}{3}\\right)\\right)+(-\\sin x\\)
\nAnswer:
\n(a) \\(\\frac{24}{25}\\)
\n(b) \\(-\\frac{24}{25}\\)
\n(c) \\(-\\frac{6}{25}\\)
\n(d) \\(\\frac{-6}{25}\\)
\nAnswer: (b)
\n\"2nd<\/p>\n

Question 5.
\nThe value of tan-1<\/sup> (tan 5)
\n(a) 2\u03c0
\n(b) 5 – 2\u03c0
\n(c) 5
\n(d) \\(\\frac{2 \\pi}{3}\\)
\nAnswer:
\n(b).
\nLet 2\u03c0 – 5 = \u03b8,
\n\u21d2 5 = 2\u03c0 – \u03b8
\ntan 5 = tan (2\u03c0 – \u03b8)
\ntan-1<\/sup> (tan 5) = tan-1<\/sup> tan (2\u03c0 – \u03b8)
\n= tan-1<\/sup> [-tan \u03b8]
\n= tan-1<\/sup> [tan (-\u03b8)] = – \u03b8
\n= – (2\u03c0 – 5) = 5 – 2\u03c0<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIf a > b > c,
\n\\(\\cot ^{-1}\\left(\\cfrac{1+a b}{a-b}\\right)+\\cot ^{-1}\\left(\\cfrac{1+b c}{b-c}\\right)+\\cot ^{-1}\\left(\\cfrac{1+a c}{c-a}\\right)\\)
\n(a) 0
\n(b) \u03c0
\n(c) 2\u03c0
\n(d) \\(\\frac{\\pi}{2}\\)
\nAnswer:
\n\"2nd<\/p>\n

2nd PUC Maths Inverse Trigonometric Functions Miscellaneous Exercise Extra Question and Answers<\/h3>\n

Question 1.
\nFind
\n\\(\\sin ^{-1} \\sin \\left(\\cfrac{3 \\pi}{5}\\right)\\) (CBSE 2010)
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nFind the principle value of \\(^{\\cot ^{-1}}\\left(\\cfrac{-1}{\\sqrt{3}}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\nWrite \\(\\cot ^{-1}\\left(\\cfrac{1}{\\sqrt{x^{2}-1}}\\right),|x|>1\\) in the simplest form
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nShow that \\(\\sin ^{-1}(2 x \\sqrt{1-x^{2}}\\}=2 \\sin ^{-1} x\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 5.
\nFind the value of tan (tan-1<\/sup> x + cot-1<\/sup> x)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 6.
\nSolve \\(\\tan ^{-1}\\left\\{\\cfrac{\\sqrt{1+x^{2}}-\\sqrt{1-x^{2}}}{\\sqrt{1+x^{2}}+\\sqrt{1-x^{2}}}\\right\\}\\) (CBSE 09)
\nAnswer:
\n\"2nd<\/p>\n

Question 7.
\nProve that \\(\\tan ^{-1} x+\\tan ^{-1} \\cfrac{2 x}{1-x^{2}}=\\tan ^{-1}\\left(\\cfrac{3 x-x^{3}}{1-3 x^{2}}\\right)\\)
\nAnswer:
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 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. Karnataka 2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Find the value …<\/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\/18086"}],"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=18086"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/18086\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=18086"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=18086"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=18086"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}