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JEE Main 2024
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Medium

Question

tan1(1+33+3)+sec1(8+436+33){\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right) is equal to :

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Solution

tan1(1+33+3)+sec1(8+436+33){\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right) =tan1(1+33+3)+sec1(16+8312+63)12= {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{16 + 8\sqrt 3 } \over {12 + 6\sqrt 3 }}} \right)^{{1 \over 2}}} =tan1(1+33(3+1))+sec1(4(12+(3)2+2.1.332(3)2+2.3.3)12 = {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {\sqrt 3 (\sqrt 3 + 1)}}} \right) + {\sec ^{ - 1}}{\left( {{{4({1^2} + {{(\sqrt 3 )}^2} + 2\,.\,1\,.\,\sqrt 3 } \over {{3^2}{{(\sqrt 3 )}^2} + 2\,.\,3\,.\,\sqrt 3 }}} \right)^{{1 \over 2}}} =tan1(13)+sec1(4(3+1)2(3+3)2)12 = {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{4{{(\sqrt 3 + 1)}^2}} \over {{{(3 + \sqrt 3 )}^2}}}} \right)^{{1 \over 2}}} =π6+sec1(2(3+1)3(3+1)) = {\pi \over 6} + {\sec ^{ - 1}}\left( {{{2(\sqrt 3 + 1)} \over {\sqrt 3 (\sqrt 3 + 1)}}} \right) =π6+sec1(23) = {\pi \over 6} + {\sec ^{ - 1}}\left( {{2 \over {\sqrt 3 }}} \right) =π6+cos1(32) = {\pi \over 6} + {\cos ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) =π6+π6 = {\pi \over 6} + {\pi \over 6} =π3 = {\pi \over 3}

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