NCERT Solutions for Class 12 Maths<\/a> Chapter 2 Inverse Trigonometric Functions Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-2-ex-2-2\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.2<\/h2>\n <\/p>\n
Ex 2.2 Class 12 NCERT Solutions Question 1.<\/strong> \n3sin-1<\/sup>x = sin-1<\/sup>(3x – 4x\u00b3). x \u2208 [-\\(\\frac { 1 }{ 2 }\\), \\(\\frac { 1 }{ 2 }\\)] \nSolution: \nLet \u03b8 = sin-1<\/sup> \u21d2 x = sin \u03b8 \n\u2234 3x – 4x\u00b3 = 3sin\u03b8 – 4sin\u00b3\u03b8 = sin3 \u03b8 \n\u2234 3\u03b8 = sin-1<\/sup>(3x – 4x\u00b3) \n\u21d2 3sin-1<\/sup>x = sin-1<\/sup>(3x – 4x\u00b3)<\/p>\nClass 12th Maths Chapter 2 Exercise 2.2 Question 2.<\/strong> \n3cos-1<\/sup>x = cos-1<\/sup>(4x\u00b3 – 3x). x \u2208 [\\(\\frac { 1 }{ 2 }\\), 1] \nSolution: \nLet \u03b8 cos-1<\/sup> \u21d2 x = cos \u03b8 \n4x\u00b3 – 3x = 4cos\u00b3\u03b8 – 3cos\u03b8 = cos 3 \u03b8 \n\u21d2 3\u03b8 = cos-1<\/sup>(4x\u00b3 – 3x) \n\u21d2 3cos-1<\/sup>x = cos-1<\/sup>(4x\u00b3 – 3x)<\/p>\nMaths Ex 2.2 Class 12 NCERT Solutions Question 3.<\/strong> \ntan-1<\/sup>\\(\\frac { 2 }{ 11 }\\) + tan-1<\/sup>\\(\\frac { 7 }{ 24 }\\) = tan-1<\/sup>\\(\\frac { 1 }{ 2 }\\) \nSolution: \n <\/p>\n <\/p>\n
Ncert Ex 2.2 Class 12\u00a0 Question 4.<\/strong> \n2tan-1<\/sup>\\(\\frac { 1 }{ 2 }\\) + tan-1<\/sup>\\(\\frac { 1 }{ 7 }\\) = tan-1<\/sup>\\(\\frac { 31 }{ 17 }\\) \nSolution: \n <\/p>\nExercise 2.2 Maths Class 12 Solutions Question 5.<\/strong> \ntan-1<\/sup>\\(\\frac{\\sqrt{1+x^{2}}-1}{x}\\), x \u2260 0 \nSolution: \n <\/p>\n <\/p>\n
Question 6. \ntan-1<\/sup>\\(\\frac{1}{\\sqrt{x^{2}-1}}\\), |x| > 1 \nSolution: \n <\/p>\nQuestion 7. \ntan-1<\/sup>\\(\\left(\\sqrt{\\frac{1-\\cos x}{1+\\cos x}}\\right)\\), x < \u03c0 \nSolution: \n <\/p>\nQuestion 8. \ntan-1<\/sup>\\(\\left(\\frac{\\cos x-\\sin x}{\\cos x+\\sin x}\\right)\\), x < \u03c0 \nSolution: \n <\/p>\nQuestion 9. \ntan-1<\/sup>\\(\\frac{x}{\\sqrt{a^{2}-x^{2}}}\\), |x| < a \nSolution: \n <\/p>\n <\/p>\n
Question 10. \ntan-1<\/sup>\\(\\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}}\\). \nSolution: \n <\/p>\nQuestion 11. \ntan-1<\/sup>[2 cos(2 sin-1<\/sup>\\(\\frac{1}{2}\\))] \nSolution: \n <\/p>\nQuestion 12. \ncot(tan-1<\/sup> a + cot-1<\/sup> a) \nSolution: \nSince tan-1<\/sup> a + cot-1<\/sup> a = \\(\\frac{\u03c0}{2}\\), \ncot(tan-1<\/sup> a + cot-1<\/sup> a = cot(\\(\\frac{\u03c0}{2}\\)) = 0<\/p>\n <\/p>\n
Question 13. \ntan\\(\\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 and xy < 1 \nSolution: \n <\/p>\n
Question 14. \nIf sin(sin-1<\/sup>\\(\\frac{1}{5}\\) + cos-1<\/sup>x) = 1, then find the value of x. \nSolution: \n <\/p>\n <\/p>\n
Question 15. \nIf tan-1<\/sup>\\(\\frac{x-1}{x-2}\\) + tan-1<\/sup>\\(\\frac{x+1}{x+2}\\), then find the value of x. \nSolution: \n <\/p>\nQuestion 16. \nsin-1<\/sup>\\(\\left(\\sin \\frac{2 \\pi}{3}\\right)\\) \nSolution: \n <\/p>\n <\/p>\n
Question 17. \ntan-1<\/sup>\\(\\left(\\tan \\frac{3 \\pi}{4}\\right)\\) \nSolution: \n <\/p>\nQuestion 18. \ntan\\(\\left(\\sin ^{-1} \\frac{3}{5}+\\cot ^{-1} \\frac{3}{2}\\right)\\) \nSolution: \n <\/p>\n
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Question 19. \ncos-1<\/sup>\\(\\left(\\cos \\frac{7 \\pi}{6}\\right)\\) is equal to \na. \\(\\frac{7\u03c0}{6}\\) \nb. \\(\\frac{5\u03c0}{6}\\) \nc. \\(\\frac{\u03c0}{3}\\) \nd. \\(\\frac{\u03c0}{6}\\) \nSolution: \nb. \\(\\frac{5\u03c0}{6}\\) \n <\/p>\nQuestion 20. \nsin\\(\\left(\\frac{\\pi}{3}-\\sin ^{-1}\\left(\\frac{-1}{2}\\right)\\right)\\) is equal to \na. \\(\\frac{1}{2}\\) \nb. \\(\\frac{1}{3}\\) \nc. \\(\\frac{1}{4}\\) \nd. 1 \nSolution: \nd. 1 \n <\/p>\n
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Question 21. \ntan-1<\/sup>\\(\\sqrt{3}\\) – cot-1<\/sup>\\(\\sqrt{3}\\) is equal to \na. \u03c0 \nb. – \\(\\frac{\u03c0}{2}\\) \nc. 0 \nd. 2\\(\\sqrt{3}\\) \nSolution: \nb. – \\(\\frac{\u03c0}{2}\\) \ntan-1<\/sup>\\(\\sqrt{3}\\) – cot-1<\/sup>(-\\(\\sqrt{3}\\)) \n= \\(\\sqrt{3}\\) – (\u03c0 – cot-1<\/sup>\\(\\sqrt{3}\\)) \n= (tan-1<\/sup>\\(\\sqrt{3}\\) + cot-1<\/sup>\\(\\sqrt{3}\\)) – \u03c0 \n= \\(\\frac{\u03c0}{2}\\) – \u03c0 = – \\(\\frac{\u03c0}{2}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-2-ex-2-2\/ NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.2 Ex 2.2 Class 12 NCERT Solutions Question 1. 3sin-1x = sin-1(3x – 4x\u00b3). x \u2208 [-, …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n