{"id":27883,"date":"2019-10-18T14:43:12","date_gmt":"2019-10-18T09:13:12","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=27883"},"modified":"2019-10-25T11:37:33","modified_gmt":"2019-10-25T06:07:33","slug":"class-12-maths-inverse-trigonometric-functions-important-questions","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/class-12-maths-inverse-trigonometric-functions-important-questions\/","title":{"rendered":"Class 12 Maths Inverse Trigonometric Functions Important Questions"},"content":{"rendered":"

Class 12 Maths Inverse Trigonometric Functions Important Questions. <\/strong>myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in\u00a0myCBSEguide<\/a>\u00a0<\/strong>website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools for years. There are around 4-5 set of solved Chapter 2 Inverse Trigonometric Functions Mathematics Extra Questions from each and every chapter. The students will not miss any concept in these Chapter wise question that are specially designed to tackle Board Exam. We have taken care of every single concept given in CBSE Class 12 Mathematics syllabus<\/a><\/strong>\u00a0and questions are framed as per the latest marking scheme and blue print issued by CBSE for class 12.<\/p>\n

Class 12 Chapter 2 Maths Extra Questions<\/strong><\/p>\n

Download as PDF<\/a><\/strong><\/p>\n

Chapter 2 Class 12 Mathematics Practice Questions<\/h2>\n

Chapter 2 Inverse Trigonometric Functions<\/strong><\/p>\n

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  1. \n
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    \n

    The period of the function f(x) = cos4x + tan3x is<\/p>\n<\/div>\n<\/div>\n

    \n
    \n
    \n
      \n
    1. {tex}\\frac{\\pi }{3}{\/tex}<\/span><\/li>\n
    2. {tex}\\pi {\/tex}<\/span><\/li>\n
    3. None of these<\/li>\n
    4. {tex}\\frac{\\pi }{2}{\/tex}<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
    5. \n
      \n
      \n

      If {tex}3{\\sin ^{ – 1}}\\left( {\\frac{{2x}}{{1 + {x^2}}}} \\right) {\/tex}<\/span>\u00a0{tex}- 4{\\cos ^{ – 1}}\\left( {\\frac{{1 – {x^2}}}{{1 + {x^2}}}} \\right) {\/tex}<\/span>{tex}+ 2{\\tan ^{ – 1}}\\left( {\\frac{{2x}}{{1 – {x^2}}}} \\right) = \\frac{\\pi }{3}{\/tex}<\/span>. Then, x=.<\/p>\n<\/div>\n<\/div>\n

      \n
      \n
      \n
        \n
      1. {tex}\\frac{1}{{\\sqrt 3 }}{\/tex}<\/span><\/li>\n
      2. {tex}\\frac{1}{{\\sqrt 2 }}{\/tex}<\/span><\/li>\n
      3. 2<\/li>\n
      4. 1<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
      5. \n
        \n
        \n

        The value of {tex}\\tan {15^0} + \\cot {15^0}{\/tex}<\/span>is<\/p>\n<\/div>\n<\/div>\n

        \n
        \n
        \n
          \n
        1. 4<\/li>\n
        2. Not defined<\/li>\n
        3. {tex}\\sqrt 3 {\/tex}<\/span><\/li>\n
        4. {tex}2\\sqrt 3 {\/tex}<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
        5. \n
          \n
          \n

          The values of x which satisfy the trigonometric equation {tex}{\\tan ^{ – 1}}\\left( {\\frac{{x – 1}}{{x – 2}}} \\right) + {\\tan ^{ – 1}}\\left( {\\frac{{x + 1}}{{x + 2}}} \\right) = \\frac{\\pi }{4}{\/tex}<\/span> are:<\/p>\n<\/div>\n<\/div>\n

          \n
          \n
          \n
            \n
          1. {tex} \\pm 2{\/tex}<\/span><\/li>\n
          2. {tex} \\pm \\frac{1}{2}{\/tex}<\/span><\/li>\n
          3. {tex} \\pm \\frac{1}{{\\sqrt 2 }}{\/tex}<\/span><\/li>\n
          4. {tex} \\pm \\sqrt 2 {\/tex}<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
          5. \n
            \n
            \n

            The minimum value of sinx – cosx is<\/p>\n<\/div>\n<\/div>\n

            \n
            \n
            \n
              \n
            1. {tex} – \\sqrt 2 {\/tex}<\/span><\/li>\n
            2. -1<\/li>\n
            3. 0<\/li>\n
            4. 1<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n
            5. The principle value of tan-1<\/sup>{tex}\\sqrt3{\/tex}<\/span> is ________.<\/li>\n
            6. If y = 2 tan-1<\/sup>x + sin-1<\/sup>{tex}\\left(\\frac{2x}{1+x^2}\\right){\/tex}<\/span> for all x, then ________ < y < ________.<\/li>\n
            7. The value of cos (sin-1<\/sup>x + cos-1<\/sup>x), |x| {tex}\\leq{\/tex}<\/span> 1 is ________.<\/li>\n
            8. \n
              \n
              \n

              Find the principal value of {tex}{\\sin ^{ – 1}}\\left( {\\frac{1}{{\\sqrt 2 }}} \\right){\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            9. \n
              \n
              \n

              Write the principal value of cos{tex}^{-1}{\/tex}<\/span>1 [cos(680)\u00b0].<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            10. \n
              \n
              \n

              Prove that {tex}{\\tan ^{ – 1}}\\sqrt x = \\frac{1}{2}{\\cos ^{ – 1}}\\left( {\\frac{{1 – x}}{{1 + x}}} \\right){\/tex}<\/span>. (<\/strong>1)<\/b><\/p>\n<\/div>\n<\/div>\n<\/li>\n

            11. \n
              \n
              \n

              Find the value of the expression {tex}{\\tan ^{ – 1}}\\left( {\\tan \\frac{{3\\pi }}{4}} \\right){\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            12. \n
              \n
              \n

              Solve the equation: 2tan-1<\/sup>(cosx) = tan-1<\/sup>(2cosec x).<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            13. \n
              \n
              \n

              Find the value of {tex}{\\sin ^{ – 1}}\\left( {\\sin \\frac{{2\\pi }}{3}} \\right){\/tex}<\/span>. (<\/strong>2)<\/b><\/p>\n<\/div>\n<\/div>\n<\/li>\n

            14. \n
              \n
              \n

              Prove that {tex} \\tan ^ { – 1 } ( 1 ) + \\tan ^ { – 1 } ( 2 ) + \\tan ^ { – 1 } ( 3 ) = \\pi.{\/tex}<\/span><\/p>\n<\/div>\n<\/div>\n<\/li>\n

            15. \n
              \n
              \n

              Solve for x, {tex} \\tan ^ { – 1 } \\frac { x } { 2 } + \\tan ^ { – 1 } \\frac { x } { 3 } = \\frac { \\pi } { 4 } , \\sqrt { 6 } > x > 0.{\/tex}<\/span><\/p>\n<\/div>\n<\/div>\n<\/li>\n

            16. \n
              \n
              \n

              Find the value of the following: {tex}{\\tan ^{ – 1}}\\left[ {2\\cos \\left( {2{{\\sin }^{ – 1}}\\frac{1}{2}} \\right)} \\right]{\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n

            17. \n
              \n
              \n

              Show that {tex}{\\sin ^{ – 1}}\\frac{{12}}{{13}} + {\\cos ^{ – 1}}\\frac{4}{5} + {\\tan ^{ – 1}}\\frac{{63}}{{16}} = \\pi{\/tex}<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n

              CBSE <\/strong>Test Paper 01<\/strong>
              \nChapter 2 Inverse Trigonometric Functions<\/strong><\/p>\n

              \n

              Solution<\/b><\/p>\n

                \n
              1. \n
                  \n
                1. {tex}\\pi {\/tex}<\/span>,\u00a0Explanation:<\/strong> {tex}f\\left( \\pi \\right) = \\left( {cos\\;4\\pi + tan3\\pi \\;} \\right) {\/tex}<\/span>\u00a0gives the same value as f (0). Therefore, the period of the function is {tex}\\pi {\/tex}<\/span>.<\/li>\n<\/ol>\n<\/li>\n
                2. \n
                    \n
                  1. {tex}\\frac{1}{{\\sqrt 3 }}{\/tex}<\/span>,\u00a0Explanation:<\/strong> {tex}3{\\sin ^{ – 1}}\\left( {\\frac{{2x}}{{1 + {x^2}}}} \\right) – 4{\\cos ^{ – 1}}\\left( {\\frac{{1 – {x^2}}}{{1 + {x^2}}}} \\right) + 2{\\tan ^{ – 1}}\\left( {\\frac{{2x}}{{1 – {x^2}}}} \\right) = \\frac{\\pi }{3}{\/tex}<\/span>
                    \nPut x = tan{tex}\\theta{\/tex}<\/span>
                    \n{tex}3{\\sin ^{ – 1}}\\left( {\\frac{{2\\tan \\theta }}{{1 + {{\\tan }^2}\\theta }}} \\right) – 4{\\cos ^{ – 1}}\\left( {\\frac{{1 – {{\\tan }^2}\\theta }}{{1 + {{\\tan }^2}\\theta }}} \\right) {\/tex}<\/span>{tex}+ 2{\\tan ^{ – 1}}\\left( {\\frac{{2\\tan \\theta }}{{1 – {{\\tan }^2}\\theta }}} \\right)\\,\\, = \\frac{\\pi }{3}{\/tex}<\/span>
                    \n{tex}3{\\sin ^{ – 1}}(\\sin 2\\theta )\\, – 4{\\cos ^{ – 1}}(\\cos 2\\theta ) + 2{\\tan ^{ – 1}}\\left( {\\tan 2\\theta } \\right)\\, = \\frac{\\pi }{3}{\/tex}<\/span>
                    \n{tex}3.2\\theta \\, – 4.2\\theta \\, + 2.2\\theta \\, = \\,\\frac{\\pi }{3}\\,\\,\\,\\, \\Rightarrow 2\\theta = \\frac{\\pi }{3}{\/tex}<\/span>{tex}\\Rightarrow \\theta = \\frac{\\pi }{6}{\/tex}<\/span>
                    \n{tex}\\therefore \\,\\,{\\tan ^{ – 1}}x = \\frac{\\pi }{6}\\,\\,\\,\\, \\Rightarrow \\,\\,x = \\tan \\left( {\\frac{\\pi }{6}} \\right)\\,\\, = \\frac{1}{{\\sqrt 3 }}{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                  2. \n
                      \n
                    1. 4,\u00a0Explanation:<\/strong> {tex}\\tan {15^0} + \\cot {15^0} = \\frac{{\\sqrt 3 – 1}}{{\\sqrt 3 + 1}} + \\frac{{\\sqrt 3 + 1}}{{\\sqrt 3 – 1}}{\/tex}<\/span>
                      \n{tex}= \\frac{{{{(\\sqrt 3 – 1)}^2} + {{(\\sqrt 3 + 1)}^2}}}{2}\\; {\/tex}<\/span>\u00a0=\u00a0{tex} \\frac{8}{2} = 4{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                    2. \n
                        \n
                      1. {tex} \\pm \\frac{1}{{\\sqrt 2 }}{\/tex}<\/span>,\u00a0Explanation:<\/strong> {tex}{\\tan ^{ – 1}}\\left( {\\frac{{x – 1}}{{x – 2}}} \\right)\\, + \\,\\,{\\tan ^{ – 1}}\\left( {\\frac{{x + 1}}{{x + 2}}} \\right)\\,\\, = \\frac{\\pi }{4}{\/tex}<\/span>
                        \n{tex}{\\tan ^{ – 1}}\\left[ {\\frac{{\\left( {\\frac{{x – 1}}{{x – 2}}} \\right)\\, + \\left( {\\frac{{x + 1}}{{x + 2}}} \\right)}}{{1 – \\left( {\\frac{{x – 1}}{{x – 2}}} \\right)\\left( {\\frac{{x + 1}}{{x + 2}}} \\right)}}} \\right]\\,\\,\\,\\,\\, = \\,\\frac{\\Pi }{4}{\/tex}<\/span>
                        \n{tex}{\\tan ^{ – 1}}\\left[ {\\frac{{\\left( {x – 1} \\right)\\,\\left( {x + 2} \\right) + \\left( {x + 1} \\right)\\left( {x – 2} \\right)}}{{\\left( {x – 2} \\right)\\left( {x + 2} \\right) – \\left( {x + 1} \\right)\\left( {x – 1} \\right)}}} \\right]\\,\\,\\, = \\,\\frac{\\Pi }{4}{\/tex}<\/span>
                        \n{tex}\\left( {\\frac{{{x^2}\\, + x – 2 + {x^2} – x – 2}}{{{x^2}\\, – 4 – {x^2} + 1}}} \\right)\\, = {\\tan ^{ – 1}}\\left( {\\frac{\\Pi }{4}} \\right){\/tex}<\/span>
                        \n{tex}\\left( {\\frac{{2{x^2} – 4}}{{ – 3}}} \\right)\\, = \\,1{\/tex}<\/span>
                        \n{tex}\\therefore \\,\\,2{x^2} – 4 = – 3{\/tex}<\/span>
                        \n{tex} \\Rightarrow \\,\\,2{x^2} = \\,1{\/tex}<\/span>
                        \n{tex}x = \\pm \\frac{1}{\\sqrt 2 }{\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                      2. \n
                          \n
                        1. {tex} – \\sqrt 2 {\/tex}<\/span>,\u00a0Explanation:<\/strong> Since, range of sine function and cosine function is [-1,1]. But, sine is increasing function and cosine is decreasing function. Therefore, the lowest that both together can attain is {tex}-45^0{\/tex}<\/span>.
                          \n{tex}\\left( { – \\frac{1}{{\\sqrt 2 }}} \\right) + \\left( { – \\frac{1}{{\\sqrt 2 }}} \\right) = – \\sqrt 2 {\/tex}<\/span><\/li>\n<\/ol>\n<\/li>\n
                        2. {tex}\\frac{\\pi}{3}{\/tex}<\/span><\/li>\n
                        3. -2{tex}\\pi{\/tex}<\/span>, 2{tex}\\pi{\/tex}<\/span><\/li>\n
                        4. 0<\/li>\n
                        5. Let {tex}{\\sin ^{ – 1}}\\left( {\\frac{1}{{\\sqrt 2 }}} \\right) = \\theta{\/tex}<\/span>
                          \n{tex}\\Rightarrow\\sin \\theta = \\frac{1}{{\\sqrt 2 }}{\/tex}<\/span>
                          \nWe know that {tex}\\theta \\in \\left[ {\\frac{{ – \\pi }}{2},\\frac{\\pi }{2}} \\right]{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\sin \\theta = \\sin \\frac{\\pi }{4}{\/tex}<\/span>\u00a0{tex}\\Rightarrow \\theta = \\frac{\\pi }{4}{\/tex}<\/span>
                          \nTherefore, principal value of {tex}{\\sin ^{ – 1}}\\left( {\\frac{1}{{\\sqrt 2 }}} \\right){\/tex}<\/span> is {tex}\\frac{\\pi }{4}{\/tex}<\/span><\/li>\n
                        6. We know that, principal value branch of {tex}\\cos ^ { – 1 }{\/tex}<\/span> x is [0, 180\u00b0].
                          \nSince, 680\u00b0 {tex}\\in{\/tex}<\/span>\u00a0[0,180\u00b0], so write 680\u00b0 as 2\u00a0{tex} \\times {\/tex}<\/span> 360\u00b0-40\u00b0
                          \nNow, {tex}\\cos ^ { – 1 }{\/tex}<\/span>[cos (680)\u00b0] = {tex}\\cos ^ { – 1 }{\/tex}<\/span> [cos(2 ;{tex} \\times{\/tex}<\/span> 360\u00b0-40\u00b0)]\n= {tex}\\cos ^ { – 1 }{\/tex}<\/span>(cos40\u00b0) {tex}[ \\because \\cos ( 4 \\pi – \\theta ) = \\cos \\theta ]{\/tex}<\/span>
                          \nSince, 40\u00b0{tex}\\in{\/tex}<\/span> [0,180\u00b0]\n{tex}\\therefore \\cos ^ { – 1 }{\/tex}<\/span>[cos(680\u00b0)] = 40\u00b0
                          \n{tex}\\left[ \\because \\cos ^ { – 1 } ( \\cos \\theta ) = \\theta ; \\forall \\theta \\in \\left[ 0,180 ^ { \\circ } \\right] \\right]{\/tex}<\/span>
                          \nwhich is the required principal value.<\/li>\n
                        7. LHS <\/strong>= {tex}{\\tan ^{ – 1}}\\sqrt x {\/tex}<\/span>
                          \nLet {tex}\\\\tan \\theta = \\sqrt x{\/tex}<\/span>
                          \n{tex}{\\tan ^2}\\theta = x{\/tex}<\/span>
                          \nR.H.S. <\/strong>{tex} = \\frac{1}{2}{\\cos ^{ – 1}}\\left( {\\frac{{1 – {{\\tan }^2}\\theta }}{{1 + {{\\tan }^2}\\theta }}} \\right){\/tex}<\/span>
                          \n{tex}= \\frac{1}{2}{\\cos ^{ – 1}}\\left( {\\cos 2\\theta } \\right) = \\frac{1}{2} \\times 2\\theta = \\theta{\/tex}<\/span>
                          \n{tex} = {\\tan ^{ – 1}}\\sqrt x{\/tex}<\/span><\/li>\n
                        8. {tex}{\\tan ^{ – 1}}\\left( {\\tan \\frac{{3\\pi }}{4}} \\right){\/tex}<\/span>
                          \n{tex} = {\\tan ^{ – 1}}\\left( {\\tan \\frac{{4\\pi – \\pi }}{4}} \\right){\/tex}<\/span>
                          \n{tex} = {\\tan ^{ – 1}}\\left[ {\\tan \\left( {\\pi – \\frac{\\pi }{4}} \\right)} \\right]{\/tex}<\/span>
                          \n{tex} = {\\tan ^{ – 1}}\\left[ { – \\tan \\frac{\\pi }{4}} \\right]{\/tex}<\/span>
                          \n{tex} = {\\tan ^{ – 1}}\\tan \\left( { – \\frac{\\pi }{4}} \\right){\/tex}<\/span>\u00a0{tex} = – \\frac{\\pi }{4}{\/tex}<\/span><\/li>\n
                        9. 2 tan-1<\/sup>(cos x) = tan-1<\/sup>(2 cosec x)
                          \n{tex}\\Rightarrow {\\tan ^{ – 1}}\\left( {\\frac{{2\\cos x}}{{1 – {{\\cos }^2}x}}} \\right) = {\\tan ^{ – 1}}\\left( {\\frac{2}{{\\sin x}}} \\right){\/tex}<\/span>
                          \n{tex}\\Rightarrow \\frac{{2\\cos x}}{{1 – {{\\cos }^2}x}} = \\frac{2}{{\\sin x}}{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\frac{{\\cos x}}{{\\sin x}} = 1{\/tex}<\/span>
                          \n{tex}\\Rightarrow{\/tex}<\/span> cot x = 1\u00a0{tex}\\Rightarrow x = \\frac{\\pi }{4}{\/tex}<\/span><\/li>\n
                        10. {tex}{\\sin ^{ – 1}}\\left( {\\sin \\frac{{2\\pi }}{3}} \\right){\/tex}<\/span>
                          \n{tex}= {\\sin ^{ – 1}}\\left( {\\sin \\frac{{3\\pi – \\pi }}{3}} \\right){\/tex}<\/span>
                          \n{tex}= {\\sin ^{ – 1}}\\left[ {\\sin \\left( {\\pi – \\frac{\\pi }{3}} \\right)} \\right]{\/tex}<\/span>
                          \n{tex}= {\\sin ^{ – 1}}\\sin \\frac{\\pi }{3}{\/tex}<\/span>\u00a0{tex} = \\frac{\\pi }{3}{\/tex}<\/span><\/li>\n
                        11. To prove,<\/strong> {tex} tan^{-1} (1) + tan^{-1} (2) + tan^{-1} (3) =\u00a0\\pi{\/tex}<\/span>
                          \nLHS<\/strong> = {tex}tan^{-1} (1) + tan^{-1} (2) + tan^{-1} (3)\u00a0{\/tex}<\/span>
                          \n{tex} = \\tan ^ { – 1 } \\left( \\tan \\frac { \\pi } { 4 } \\right) + \\frac { \\pi } { 2 } – \\cot ^ { – 1 } ( 2 ) + \\frac { \\pi } { 2 } – \\cot ^ { – 1 } ( 3 ){\/tex}<\/span> {tex} \\left[ \\because \\tan ^ { – 1 } x + \\cot ^ { – 1 } x = \\frac { \\pi } { 2 } \\right]{\/tex}<\/span>
                          \n{tex} = \\frac { \\pi } { 4 } + \\pi – \\left[ \\cot ^ { – 1 } ( 2 ) + \\cot ^ { – 1 } ( 3 ) \\right]{\/tex}<\/span>{tex} \\left[ \\because \\tan ^ { – 1 } ( \\tan \\theta ) = \\theta ; \\forall \\theta \\in \\left( – \\frac { \\pi } { 2 } , \\frac { \\pi } { 2 } \\right) \\right]{\/tex}<\/span>
                          \n{tex} = \\frac { 5 \\pi } { 4 } – \\left[ \\tan ^ { – 1 } \\left( \\frac { 1 } { 2 } \\right) + \\tan ^ { – 1 } \\left( \\frac { 1 } { 3 } \\right) \\right]{\/tex}<\/span>{tex} \\left[ \\because \\cot ^ { – 1 } x = \\tan ^ { – 1 } \\frac { 1 } { x } , x > 0 \\right]{\/tex}<\/span>
                          \n{tex} = \\frac { 5 \\pi } { 4 } – \\left[ \\tan ^ { – 1 } \\left( \\frac { \\frac { 1 } { 2 } + \\frac { 1 } { 3 } } { 1 – \\frac { 1 } { 2 } \\cdot \\frac { 1 } { 3 } } \\right) \\right]{\/tex}<\/span>{tex} \\left[ \\because \\tan ^ { – 1 } x + \\tan ^ { – 1 } y = \\tan ^ { – 1 } \\left( \\frac { x + y } { 1 – x y } \\right), \\text { if } x y < 1 \\right]{\/tex}<\/span>
                          \n{tex} = \\frac { 5 \\pi } { 4 } – \\tan ^ { – 1 } \\left( \\frac { 5 \/ 6 } { 5 \/ 6 } \\right){\/tex}<\/span>
                          \n{tex} = \\frac { 5 \\pi } { 4 } – \\tan ^ { – 1 } ( 1 ) = \\frac { 5 \\pi } { 4 } – \\frac { \\pi } { 4 } = \\frac { 4 \\pi } { 4 } = \\pi{\/tex}<\/span>= RHS (<\/strong>Hence Proved)<\/strong><\/li>\n
                        12. Here, we have to find the value of x .Now, we are given that
                          \n{tex} \\tan ^ { – 1 } \\frac { x } { 2 } + \\tan ^ { – 1 } \\frac { x } { 3 } {\/tex}<\/span>{tex}= \\frac { \\pi } { 4 } , \\sqrt { 6 } > x > 0 {\/tex}<\/span>
                          \n{tex}\\Rightarrow \\tan ^ { – 1 } \\left( \\frac { \\frac { x } { 2 } + \\frac { x } { 3 } } { 1 – \\frac { x ^ { 2 } } { 6 } } \\right) = \\frac { \\pi } { 4 }{\/tex}<\/span>{tex} \\left[ \\because \\tan ^ { – 1 } x + \\tan ^ { – 1 } y = \\tan ^ { – 1 } \\left( \\frac { x + y } { 1 – x y } \\right) ; x y < 1 \\right]{\/tex}<\/span>
                          \n{tex} \\Rightarrow \\quad \\frac { \\frac { 3 x + 2 x } { 6 } } { \\frac { 6 – x ^ { 2 } } { 6 } } = \\tan \\frac { \\pi } { 4 }{\/tex}<\/span> { taking tan on both sides}
                          \n{tex}\\Rightarrow \\frac { 5 x } { 6 – x ^ { 2 } } = 1 \\left[ \\because \\tan \\frac { \\pi } { 4 } = 1 \\right]{\/tex}<\/span>
                          \n{tex} \\Rightarrow{\/tex}<\/span> 5x = 6-x2<\/sup>
                          \n{tex}\\Rightarrow{\/tex}<\/span> x2<\/sup> + 5x – 6 = 0
                          \n{tex}\\Rightarrow{\/tex}<\/span> x2<\/sup> + 6x – x – 6 = 0
                          \n{tex}\\Rightarrow{\/tex}<\/span> x (x + 6) – 1 (x + 6) = 0
                          \n{tex}\\Rightarrow{\/tex}<\/span> (x-1) (x + 6) = 0
                          \n{tex}\\therefore{\/tex}<\/span> x = 1 or – 6
                          \nBut it is given that, {tex} \\,\\sqrt {\\text{6}} {\\text{ > x > 0 }} \\Rightarrow {\\text{x > 0}}{\/tex}<\/span>
                          \n{tex} \\therefore{\/tex}<\/span> x = – 6 is rejected.
                          \nHence, x = 1 is the only solution of the given equation.<\/li>\n
                        13. {tex}{\\tan ^{ – 1}}\\left[ {2\\cos \\left( {2{{\\sin }^{ – 1}}\\frac{1}{2}} \\right)} \\right]{\/tex}<\/span>
                          \n{tex} = {\\tan ^{ – 1}}\\left[ {2\\cos \\left( {2{{\\sin }^{ – 1}}\\sin \\frac{\\pi }{6}} \\right)} \\right]{\/tex}<\/span>
                          \n{tex} = {\\tan ^{ – 1}}\\left[ {2\\cos \\left( {2 \\times \\frac{\\pi }{6}} \\right)} \\right]{\/tex}<\/span>
                          \n{tex} = {\\tan ^{ – 1}}\\left[ {2\\cos \\frac{\\pi }{3}} \\right]{\/tex}<\/span>
                          \n{tex}= {\\tan ^{ – 1}}\\left[ {2 \\times \\frac{1}{2}} \\right]{\/tex}<\/span>\u00a0= tan-1<\/sup>1
                          \n{tex}= {\\tan ^{ – 1}}\\tan \\frac{\\pi }{4} = \\frac{\\pi }{4}{\/tex}<\/span><\/li>\n
                        14. Let {tex}\\theta {\/tex}<\/span> = sin-1<\/sup>({tex}\\frac{12}{13}{\/tex}<\/span>)
                          \n{tex}\\Rightarrow{\/tex}<\/span> sin{tex}\\theta{\/tex}<\/span> = {tex}\\frac{12}{13}{\/tex}<\/span>
                          \n{tex}\\Rightarrow\\sqrt{1-cos^2\\theta}=\\frac{12}{13}{\/tex}<\/span>
                          \n{tex}\\Rightarrow 1-cos^2\\theta=\\frac{(12)^2}{(13)^2}{\/tex}<\/span>
                          \n{tex}\\Rightarrow cos^2\\theta=\\frac{(5)^2}{(13)^2}{\/tex}<\/span>
                          \n{tex}\\Rightarrow cos\\theta=\\frac{5}{13}{\/tex}<\/span>
                          \nSince, tan{tex}\\theta{\/tex}<\/span> = {tex}\\frac{sin\\theta}{cos\\theta}=\\frac{\\frac{12}{13}}{\\frac{5}{13}}=\\frac{12}{5}{\/tex}<\/span>
                          \n{tex}\\Rightarrow \\theta=tan^{-1}(\\frac{12}{5}){\/tex}<\/span>
                          \nThus, {tex}\\theta=sin^{-1}(\\frac{12}{13})=tan^{-1}(\\frac{12}{5}){\/tex}<\/span>
                          \nSimilarly, {tex}cos^{-1}(\\frac{5}{13})=tan^{-1}(\\frac{12}{5}){\/tex}<\/span>
                          \nWe have, LHS =\u00a0{tex}{\\sin ^{ – 1}}\\left( {\\frac{{12}}{{13}}} \\right) + {\\cos ^{ – 1}}\\left( {\\frac{4}{5}} \\right) + {\\tan ^{ – 1}}\\left( {\\frac{{63}}{{16}}} \\right){\/tex}<\/span>
                          \n{tex} = {\\tan ^{ – 1}}\\left( {\\frac{{12}}{5}} \\right) + {\\tan ^{ – 1}}\\left( {\\frac{3}{4}} \\right) + {\\tan ^{ – 1}}\\left( {\\frac{{63}}{{16}}} \\right){\/tex}<\/span>
                          \n{tex} = \\left[ {{{\\tan }^{ – 1}}\\left( {\\frac{{12}}{5}} \\right) + {{\\tan }^{ – 1}}\\left( {\\frac{3}{4}} \\right)} \\right] + {\\tan ^{ – 1}}\\left( {\\frac{{63}}{{16}}} \\right){\/tex}<\/span>
                          \n{since {tex}\\frac{{12}}{5} \\times \\frac{3}{4} = \\frac{9}{5} > 1,{\/tex}<\/span> therefore , tan-1<\/sup>A + tan-1<\/sup>B =\u00a0{tex}\\pi + {\\tan ^{ – 1}}\\frac{{A + \\_B}}{{1 – AB}}{\/tex}<\/span>)
                          \n={tex}\\pi + {\\tan ^{ – 1}}\\left( {\\frac{{\\frac{{12}}{5} + \\frac{3}{4}}}{{1 – \\left( {\\frac{{12}}{5}} \\right)\\left( {\\frac{3}{4}} \\right)}}} \\right) + {\\tan ^{ – 1}}\\left( {\\frac{{63}}{{16}}} \\right){\/tex}<\/span>
                          \n= {tex}\\pi + {\\tan ^{ – 1}}\\left( { – \\frac{{63}}{{16}}} \\right) + {\\tan ^{ – 1}}\\left( {\\frac{{63}}{{16}}} \\right){\/tex}<\/span>
                          \n{tex} = \\pi – {\\tan ^{ – 1}}\\left( {\\frac{{63}}{{16}}} \\right) + {\\tan ^{ – 1}}\\left( {\\frac{{63}}{{16}}} \\right){\/tex}<\/span>\u00a0= {tex}\\pi{\/tex}<\/span>\u00a0Hence Proved.<\/li>\n<\/ol>\n

                          Chapter Wise Important Questions Class 12 Maths Part I and Part II<\/h2>\n
                            \n
                          1. Relations and Functions<\/a><\/li>\n
                          2. Inverse Trigonometric Functions<\/a><\/li>\n
                          3. Matrices<\/a><\/li>\n
                          4. Determinants<\/a><\/li>\n
                          5. Continuity and Differentiability<\/a><\/li>\n
                          6. Application of Derivatives<\/a><\/li>\n
                          7. Integrals<\/a><\/li>\n
                          8. Application of Integrals<\/a><\/li>\n
                          9. Differential Equations<\/a><\/li>\n
                          10. Vector Algebra<\/a><\/li>\n
                          11. Three Dimensional Geometry<\/a><\/li>\n
                          12. Linear Programming<\/a><\/li>\n
                          13. Probability<\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                            Class 12 Maths Inverse Trigonometric Functions Important Questions. myCBSEguide has just released Chapter Wise Question Answers for class 12 Maths. There chapter wise Practice Questions with complete solutions are available for download in\u00a0myCBSEguide\u00a0website and mobile app. These Questions with solution are prepared by our team of expert teachers who are teaching grade in CBSE schools … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1346,1432],"tags":[1867,1839,1838,1833,1832,1854],"class_list":["post-27883","post","type-post","status-publish","format-standard","hentry","category-cbse","category-mathematics-cbse-class-12","tag-cbse-class-12-mathematics","tag-extra-questions","tag-important-questions","tag-latest-exam-questions","tag-practice-questions","tag-practice-test"],"yoast_head":"\nClass 12 Maths Inverse Trigonometric Functions Important Questions<\/title>\n<meta name=\"description\" content=\"Class 12 Maths Inverse Trigonometric Functions Important 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