{"id":169997,"date":"2021-02-25T06:45:12","date_gmt":"2021-02-25T06:45:12","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=169997"},"modified":"2023-09-21T07:21:51","modified_gmt":"2023-09-21T07:21:51","slug":"ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/","title":{"rendered":"NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions"},"content":{"rendered":"\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Class 12: Maths Chapter 2 solutions. Complete Class 12 <meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Maths Chapter 2 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\"><strong>NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions<\/strong><\/h2>\n\n\n\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Class 12: <meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Maths Chapter 2 solutions. Complete Class 12 <meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Maths Chapter 2 Notes.<\/p>\n\n\n\n<p>Page No: 41<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-exercise-2-1\">Exercise 2.1<\/h2>\n\n\n\n<p><strong>Find the principal values of the following:<br>1. sin<sup>-1<\/sup>(-1\/2)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>1. Let sin<sup>-1<\/sup>(\u22121\/2) = y, then<\/p>\n\n\n\n<p>sin y = \u22121\/2 = \u2212sin(\u03c0\/6) = sin(\u2212\u03c0\/6)<\/p>\n\n\n\n<p>Range of the principal value of sn<sup>-1<\/sup> is [-\u03c0\/2, \u03c0\/2] and sin -\u03c0\/6) = -1\/2<\/p>\n\n\n\n<p>Therefore, the principal value of sin<sup>-1<\/sup>(-1\/2) is -\u03c0\/6.<\/p>\n\n\n\n<p><strong>2. cos<sup>-1<\/sup>(\u221a3\/2)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let cos<sup>-1<\/sup>(\u221a3\/2) = y,<\/p>\n\n\n\n<p>cos y = \u221a3\/2 = cos (\u03c0\/6)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of cos<sup>-1<\/sup>&nbsp;is [0, \u03c0] and cos (\u03c0\/6) = \u221a3\/2<\/p>\n\n\n\n<p>Therefore, the principal value of cos<sup>-1<\/sup>(\u221a3\/2) is \u03c0\/6.<\/p>\n\n\n\n<p><strong>3. cosec<sup>-1<\/sup>(2)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let cosec<sup>-1<\/sup>(2) = y.<\/p>\n\n\n\n<p>Then, cosec y = 2 = cosec (\u03c0\/6)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of cosec<sup>-1<\/sup>&nbsp;is [-\u03c0\/2, \u03c0\/2] &#8211; {0} and cosec (\u03c0\/6) = 2.<\/p>\n\n\n\n<p>Therefore, the principal value of cosec<sup>-1<\/sup>(2) is \u03c0\/6.<\/p>\n\n\n\n<p><strong>4. tan<sup>-1<\/sup>(\u221a3)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let tan<sup>-1<\/sup>(-\u221a3) = y,<\/p>\n\n\n\n<p>then tan y = -\u221a3 = -tan \u03c0\/3 = tan (-\u03c0\/3)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of tan<sup>-1<\/sup>&nbsp;is (-\u03c0\/2, \u03c0\/2) and tan (-\u03c0\/3)<\/p>\n\n\n\n<p>= -\u221a3<\/p>\n\n\n\n<p>Therefore, the principal value of tan<sup>-1<\/sup>&nbsp;(-\u221a3) is -\u03c0\/3<\/p>\n\n\n\n<p><strong>5. cos<sup>-1<\/sup>(-1\/2)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let cos<sup>-1<\/sup>(-1\/2) = y,<\/p>\n\n\n\n<p>then cos y = -1\/2 = -cos \u03c0\/3 = cos (\u03c0-\u03c0\/3) = cos (2\u03c0\/3)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of cos<sup>-1<\/sup>&nbsp;is [0, \u03c0] and cos (2\u03c0\/3) = -1\/2<\/p>\n\n\n\n<p>Therefore, the principal value of cos<sup>-1<\/sup>(-1\/2) is 2\u03c0<\/p>\n\n\n\n<p><strong>6. tan<sup>-1<\/sup>(-1)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let tan<sup>-1<\/sup>(-1) = y. Then, tan y = -1 = -tan (\u03c0\/4) = tan (-\u03c0\/4)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of tan<sup>-1<\/sup>&nbsp;is (-\u03c0\/2, \u03c0\/2) and tan (-\u03c0\/4) = -1.<\/p>\n\n\n\n<p>Therefore, the principal value of tan<sup>-1<\/sup>(\u22121) is -\u03c0\/4.<\/p>\n\n\n\n<p><strong>7. sec<sup>-1<\/sup>(2\/\u221a3)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let sec<sup>-1<\/sup>(2\/\u221a3) = y, then sec y = 2\/\u221a3 = sec (\u03c0\/6)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of sec<sup>-1<\/sup>&nbsp;is [0, \u03c0] \u2212 {\u03c0\/2} and sec (\u03c0\/6) = 2\/\u221a3.<\/p>\n\n\n\n<p>Therefore, the principal value of sec<sup>-1<\/sup>(2\/\u221a3) is \u03c0\/6.<\/p>\n\n\n\n<p><strong>8. cot<sup>-1<\/sup>(\u221a3)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let cot<sup>-1<\/sup>\u221a3 = y, then cot y = \u221a3 = cot (\u03c0\/6).<\/p>\n\n\n\n<p>We know that the range of the principal value branch of cot<sup>-1<\/sup>&nbsp;is (0, \u03c0) and cot (\u03c0\/6) = \u221a3.<\/p>\n\n\n\n<p>Therefore, the principal value of cot<sup>-1<\/sup>\u221a3 is \u03c0.<\/p>\n\n\n\n<p><strong>9. cos<sup>-1<\/sup>(-1\/\u221a2)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let cos<sup>-1<\/sup>(-1\/\u221a2) = y,<\/p>\n\n\n\n<p>then cos y = -1\/\u221a2 = -cos (\u03c0\/4) = cos (\u03c0 &#8211; \u03c0\/4) = cos (3\u03c0\/4).<\/p>\n\n\n\n<p>We know that the range of the principal value branch of cos<sup>-1<\/sup>&nbsp;is [0, \u03c0] and cos (3\u03c04) = -1\/\u221a2.<\/p>\n\n\n\n<p>Therefore, the principal value of cos<sup>-1<\/sup>(-1\/\u221a2) is 3\u03c0\/4.<\/p>\n\n\n\n<p><strong>10. cosec<sup>-1<\/sup>(-\u221a2)&nbsp;&nbsp;<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let cosec<sup>-1<\/sup>(\u2212\u221a2) = y, then cosec y = \u2212\u221a2 = \u2212cosec (\u03c0\/4) = cosec (\u2212\u03c0\/4)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of cosec<sup>-1<\/sup>&nbsp;is [-\u03c0\/2, \u03c0\/2]-{0} and cosec(-\u03c0\/4) = -\u221a2.<\/p>\n\n\n\n<p>Therefore, the principal value of cosecc<sup>-1<\/sup>(-\u221a2) is -\u03c0\/4.<\/p>\n\n\n\n<p>Page No. 42<\/p>\n\n\n\n<p><strong>Find the values of the following:<\/strong><\/p>\n\n\n\n<p><strong>11. tan<sup>-1<\/sup>(1) + cos<sup>-1<\/sup>(-1\/2) + sin<sup>-1<\/sup>(-1\/2)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let tan<sup>-1<\/sup>(1) = x,<\/p>\n\n\n\n<p>then tan x = 1 = tan(\u03c0\/4)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of tan<sup>-1<\/sup>&nbsp;is (\u2212\u03c0\/2, \u03c0\/2).<\/p>\n\n\n\n<p>\u2234 tan<sup>-1<\/sup>(1) = \u03c0\/4<\/p>\n\n\n\n<p>Let cos<sup>-1<\/sup>(\u22121\/2) = y,<\/p>\n\n\n\n<p>then cos y = \u22121\/2 = \u2212cos\u03c0\/3 = cos (\u03c0 \u2212 \u03c0\/3)<\/p>\n\n\n\n<p>= cos (2\u03c0\/3)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of cos<sup>-1<\/sup>&nbsp;is [0, \u03c0].<\/p>\n\n\n\n<p>\u2234 cos<sup>-1<\/sup>(\u22121\/2) = 2\u03c0\/3<\/p>\n\n\n\n<p>Let sin<sup>-1<\/sup>(\u22121\/2) = z,<\/p>\n\n\n\n<p>then sin z = \u22121\/2 = \u2212sin \u03c0\/6 = sin (\u2212\u03c0\/6)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of sin<sup>-1<\/sup>&nbsp;is [-\/\u03c02, \u03c0\/2].<\/p>\n\n\n\n<p>\u2234 sin<sup>-1<\/sup>(-1\/2) = -\u03c0\/6<\/p>\n\n\n\n<p>Now,<\/p>\n\n\n\n<p>tan<sup>-1<\/sup>(1) + cos<sup>-1<\/sup>(-1\/2) + sin<sup>-1<\/sup>(-1\/2)<\/p>\n\n\n\n<p>= \u03c0\/4 + 2\u03c0\/3 \u2212 \u03c0\/6<\/p>\n\n\n\n<p>= (3\u03c0 + 8\u03c0 \u2212 2\u03c0)\/12<\/p>\n\n\n\n<p>= 9\u03c0\/12 = 3\u03c0\/4<\/p>\n\n\n\n<p><strong>12. cos<sup>-1<\/sup>(1\/2) + 2 sin<sup>-1<\/sup>(1\/2)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let cos<sup>-1<\/sup>(1\/2) = x, then<br>cos x = 1\/2 = cos \u03c0\/3<br>We know that the range of the principal value branch of cos\u22121 is [0, \u03c0].<br>\u2234 cos<sup>-1<\/sup>(1\/2)<br>= \u03c0\/3<br>Let sin<sup>-1<\/sup>(-1\/2) = y, then<br>sin y = 1\/2<br>= sin \u03c0\/6<\/p>\n\n\n\n<p>We know that the range of the principal value branch of sin<sup>-1<\/sup>&nbsp;is [-\u03c0\/2, \u03c0\/2].<br>\u2234 sin<sup>-1<\/sup>(1\/2) = \u03c0\/6<\/p>\n\n\n\n<p>Now,<br>cos<sup>-1<\/sup>(1\/2) + 2sin<sup>-1<\/sup>(1\/2)<br>= \u03c0\/3 + 2\u00d7\u03c0\/6<br>= \u03c0\/3 + \u03c0\/3<br>= 2\u03c0\/3<\/p>\n\n\n\n<p><strong>13. If sin<sup>-1<\/sup>&nbsp;x = y, then<\/strong><\/p>\n\n\n\n<p>(A) 0 \u2264 y \u2264 \u03c0<\/p>\n\n\n\n<p>(B) -\u03c0\/2 \u2264 y \u2264 \u03c0\/2<\/p>\n\n\n\n<p>(C) 0 &lt; y &lt; \u03c0&nbsp;<\/p>\n\n\n\n<p>(D) -\u03c0\/2 &lt; y &lt; \u03c0\/2<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>It is given that sin<sup>-1<\/sup>x = y.<\/p>\n\n\n\n<p>We know that the range of the principal value branch of sin<sup>-1<\/sup>&nbsp;is [-\u03c0\/2, \u03c0\/2].<\/p>\n\n\n\n<p>Therefore, -\u03c0\/2 \u2264 y \u2264 \u03c0\/2.<\/p>\n\n\n\n<p>Hence, the option (B) is correct.<\/p>\n\n\n\n<p><strong>14. tan<sup>-1<\/sup>\u221a3 &#8211; sec<sup>-1<\/sup>(-2) is equal to<\/strong><\/p>\n\n\n\n<p>(A) \u03c0<\/p>\n\n\n\n<p>(B) -\u03c0\/3<\/p>\n\n\n\n<p>(C) \u03c0\/3<\/p>\n\n\n\n<p>(D) 2\u03c0\/3<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>Let tan<sup>-1<\/sup>\u221a3 = x,then<\/p>\n\n\n\n<p>tan x = \u221a3 = tan \u03c0\/3<\/p>\n\n\n\n<p>We know that the range of the principal value branch of tan<sup>-1<\/sup>&nbsp;is (-\u03c0\/2, \u03c0\/2).<\/p>\n\n\n\n<p>\u2234 tan<sup>-1<\/sup>\u221a3 = \u03c0\/3<\/p>\n\n\n\n<p>Let sec<sup>-1<\/sup>(-2) = y, then<\/p>\n\n\n\n<p>sec y = -2 = -sec \u03c0\/3<\/p>\n\n\n\n<p>= sec (\u03c0 &#8211; \u03c0\/3)<\/p>\n\n\n\n<p>= sec (2\u03c0\/3)<\/p>\n\n\n\n<p>We know that the range of the principal value branch of sec<sup>-1<\/sup>&nbsp;is [0, \u03c0]- {\u03c0\/2}<\/p>\n\n\n\n<p>\u2234 sec<sup>-1<\/sup>(-2) =2\u03c0\/3<\/p>\n\n\n\n<p>Now,<\/p>\n\n\n\n<p>tan<sup>-1<\/sup>\u221a3 &#8211; sec<sup>-1<\/sup>(-2)<\/p>\n\n\n\n<p>= \u03c0\/3 &#8211; 2\u03c0\/3<\/p>\n\n\n\n<p>= -\u03c0\/3<\/p>\n\n\n\n<p>Hence, the option (B) is correct.<\/p>\n\n\n\n<p>Page No: 41<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-exercise-2-1-0\">Exercise 2.1<\/h2>\n\n\n\n<p><strong>Prove the following:<br>1. 3sin<sup>-1<\/sup>x = sin<sup>-1<\/sup>(3x \u2013 4&#215;3), x \u2208 [-\/2, 1\/2]<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>To prove:<\/p>\n\n\n\n<p>3sin<sup>-1<\/sup>x = sin<sup>-1<\/sup>(3x \u2212 4&#215;3), x \u2208 [\u22121\/2, 1\/2]<\/p>\n\n\n\n<p>Let sin<sup>-1<\/sup>x = \u03b8, then x = sin \u03b8.<\/p>\n\n\n\n<p>We have,<\/p>\n\n\n\n<p>RHS = sin-1(3x &#8211; 4x<sup>3<\/sup>)<\/p>\n\n\n\n<p>= sin<sup>-1<\/sup>(3 sin \u03b8 &#8211; 4sin3\u03b8)<\/p>\n\n\n\n<p>= sin<sup>-1<\/sup>(sin 3\u03b8) = 3\u03b8<\/p>\n\n\n\n<p>= 3sin<sup>-1<\/sup>x = LHS<\/p>\n\n\n\n<p><strong>2. 3cos<sup>-1<\/sup>x = cos<sup>-1<\/sup>(4x<sup>3<\/sup>&nbsp;\u2013 3x) x \u2208 [1, 1\/2]<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>To prove:<\/p>\n\n\n\n<p>3cos<sup>-1<\/sup>x = cos<sup>-1<\/sup>(4x<sup>3<\/sup>&nbsp;\u2013 3x) x \u2208 [1, 1\/2].<\/p>\n\n\n\n<p>Let cos<sup>-1<\/sup>x = \u03b8, then x = cos \u03b8.<\/p>\n\n\n\n<p>We have,<\/p>\n\n\n\n<p>RHS = cos<sup>-1<\/sup>(4x<sup>3<\/sup>&nbsp;&#8211; 3x)<\/p>\n\n\n\n<p>= cos<sup>-1<\/sup>(4cos<sup>3<\/sup>\u03b8 &#8211; 3cos\u03b8)<\/p>\n\n\n\n<p>= cos<sup>-1<\/sup>(cos 3\u03b8) = 3\u03b8<\/p>\n\n\n\n<p>= 3cos<sup>-1<\/sup>x<\/p>\n\n\n\n<p>= LHS<\/p>\n\n\n\n<p><strong>3. tan<sup>-1<\/sup>2\/11 + tan<sup>-1<\/sup>7\/24 = tan<sup>-1<\/sup>1\/2<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>To prove: tan<sup>-1<\/sup>2\/11 + tan<sup>-1<\/sup>7\/24 = tan<sup>-1<\/sup>1\/2<br>LHS = tan<sup>-1<\/sup>2\/11 + tan<sup>-1<\/sup>7\/24<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-1.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p>= tan<sup>-1<\/sup>(48 + 77)\/(264 \u2212 14)<\/p>\n\n\n\n<p>= tan<sup>-1<\/sup>125\/250 = tan<sup>-1<\/sup>1\/2 = RHS<\/p>\n\n\n\n<p>4. 2tan<sup>-1<\/sup>1\/2 + tan<sup>-1<\/sup>1\/7 = tan<sup>-1<\/sup>31\/17<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>To prove: 2tan<sup>-1<\/sup>1\/2 + tan<sup>-1<\/sup>1\/7 = tan<sup>-1<\/sup>31\/17<\/p>\n\n\n\n<p>LHS = 2tan-11\/2 + tan-11\/7<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-2.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p>Write the following functions in the simplest form:<\/p>\n\n\n\n<p><strong>Question: 5<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-3.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-4.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 6<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-5.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-6.jpgG\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 7<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-7.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-8.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 8<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-9.jpg\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-10.png\" alt=\"\"\/><\/figure>\n\n\n\n<p>Page No. 48<\/p>\n\n\n\n<p><strong>Question: 9<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-11.png\" alt=\"\"\/><\/figure>\n\n\n\n<p>Answer<\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-12.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 10<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-13.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-14.png\" alt=\"\"\/><\/figure>\n\n\n\n<p>Find the values of each of the following:<\/p>\n\n\n\n<p><strong>Question: 11<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-15.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-16.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 12. cot (tan<sup>-1<\/sup>a + cot<sup>-1<\/sup>a)<\/strong><\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>The given function is cot(tan<sup>-1<\/sup>a + cot<sup>-1<\/sup>a).<\/p>\n\n\n\n<p>\u2234 cot(tan<sup>-1<\/sup>a + cot<sup>-1<\/sup>a)<\/p>\n\n\n\n<p>= cot (\u03c0\/2) [tan<sup>-1<\/sup>x + cot<sup>-1<\/sup>x = \u03c0\/2]<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p><strong>Question: 13<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-17.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-18.png\" alt=\"\"\/><\/figure>\n\n\n\n<p>Formula used:<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"365\" height=\"60\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/mbnmnb.png\" alt=\"\" class=\"wp-image-170162\" srcset=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/mbnmnb.png 365w, https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/mbnmnb-300x49.png 300w\" sizes=\"auto, (max-width: 365px) 100vw, 365px\" \/><\/figure>\n\n\n\n<p><strong>Question: 14<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-19.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-20.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 15<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-21.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-22.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 16<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-23.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-24.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 17<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-25.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-26.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 18<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-27.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-28.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Question: 19<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-29.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\" alt=\"\"\/><\/figure>\n\n\n\n<p>The correct option is B.<\/p>\n\n\n\n<p><strong>Question: 20<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-31.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-32.png\" alt=\"\"\/><\/figure>\n\n\n\n<p>The correct option is D.<\/p>\n\n\n\n<p><strong>Question: 21<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-33.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image\"><img decoding=\"async\" src=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/class12-math-inverse-trigonometric-functions-ncert-solutions-34.png\" alt=\"\"\/><\/figure>\n\n\n\n<p>The correct option is B.<\/p>\n\n\n\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Class 12: <meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Maths Chapter 2 solutions. Complete Class 12 <meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Maths Chapter 2 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-solutions-for-12th-class-maths-chapter-2-nbsp-download-pdf\">NCERT Solutions for 12th Class Maths: Chapter 2:&nbsp;<strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/NCERT-Solutions-for-12th-Class-Maths_-Chapter-2-Inverse-Trigonometric-Functions.pdf\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: <meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Chapterwise NCERT Solutions for Class 12 Maths :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-1-relations-and-functions\/\">Chapter 1 \u2013 Relations and Functions<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/\">Chapter 2 \u2013 Inverse Trigonometric Functions.<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-3-matrices\/\">Chapter 3 \u2013 Matrices<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-4-determinants\/\">Chapter 4 \u2013 Determinants.<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-5-continuity-and-differentiability\/\">Chapter 5 \u2013 Continuity and Differentiability.0.0<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-6-application-of-derivatives\/\">Chapter 6 \u2013 Application of Derivatives.<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-7-integrals\/\">Chapter 7 \u2013 Integrals.<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-8-application-of-integrals\/\">Chapter 8 \u2013 Application of Integrals.<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-9-differential-equations\/\">Chapter 9: Differential Equations<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-10-vector-algebra\/\">Chapter 10: Vector Algebra<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-11-three-dimensional-geometry\/\">Chapter 11: Three Dimensional Geometry<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-12-linear-programming\/\">Chapter 12: Linear Programming<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-13-probability\/\">Chapter 13: Probability<\/a><\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-about-ncert\">About NCERT<\/h2>\n\n\n\n<p>The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. <a href=\"https:\/\/ncert.nic.in\/\" target=\"_blank\" rel=\"noreferrer noopener\">Visit the Official NCERT website<\/a> to learn more. <\/p>\n\n\n\n<div class=\"wp-block-buttons is-layout-flex wp-block-buttons-is-layout-flex\">\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions\/\">NCERT Solutions<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-class-xii\/\">NCERT Solutions for Class 12<\/a><\/div>\n\n\n\n<div class=\"wp-block-button\"><a class=\"wp-block-button__link has-primary-background-color has-background wp-element-button\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-12-maths\/\">NCERT Solutions for Class 12 <strong>Maths<\/strong><\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Class 12: Maths Chapter 2 solutions. Complete Class 12 Maths Chapter 2 Notes. NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions Class 12: Maths Chapter 2 solutions. Complete Class 12 Maths Chapter 2 Notes. Page No: 41 Exercise 2.1 Find the principal values of the following:1. sin-1(-1\/2) Answer 1. Let sin-1(\u22121\/2) = y, [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":628474,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,25],"tags":[1557],"boards":[1180],"class_list":["post-169997","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-12","tag-ncert-maths-class-12","boards-ncert","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions for Class 12, Maths Chapter 2 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions | Browse Class 12 Maths Chapters NCERT - IndCareer Schools\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions\" \/>\n<meta property=\"og:description\" content=\"Class 12: Maths Chapter 2 solutions. Complete Class 12 Maths Chapter 2 Notes. NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2021-02-25T06:45:12+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-21T07:21:51+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/NCERT-Solutions-47-3-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"900\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"14 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"NCERT Solutions for 12th Class Maths: Chapter 2-Inverse Trigonometric Functions\",\"datePublished\":\"2021-02-25T06:45:12+00:00\",\"dateModified\":\"2023-09-21T07:21:51+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/\"},\"wordCount\":1046,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/02\/NCERT-Solutions-47-3-scaled.jpg\",\"keywords\":[\"NCERT Maths (Class 12)\"],\"articleSection\":[\"Book Solutions\",\"Class 12\"],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/\",\"url\":\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-12th-class-maths-chapter-2-inverse-trigonometric-functions\/\",\"name\":\"NCERT Solutions for Class 12, Maths Chapter 2 - 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