^-1(1+√33+√3)+^-1(8+4√36+3√3) is equal to

Question

$	an^{-1}\left(\dfrac{1+\sqrt3}{3+\sqrt3}ight)+\sec^{-1}\left(\dfrac{8+4\sqrt3}{6+3\sqrt3}ight)$ is equal to

Accepted Answer

$${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$$

$$= {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{16 + 8\sqrt 3 } \over {12 + 6\sqrt 3 }}} \right)^{{1 \over 2}}}$$

$$ = {\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {\sqrt 3 (\sqrt 3  + 1)}}} \right) + {\sec ^{ - 1}}{\left( {{{4({1^2} + {{(\sqrt 3 )}^2} + 2\,.\,1\,.\,\sqrt 3 } \over {{3^2}{{(\sqrt 3 )}^2} + 2\,.\,3\,.\,\sqrt 3 }}} \right)^{{1 \over 2}}}$$

$$ = {\tan ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right) + {\sec ^{ - 1}}{\left( {{{4{{(\sqrt 3  + 1)}^2}} \over {{{(3 + \sqrt 3 )}^2}}}} \right)^{{1 \over 2}}}$$

$$ = {\pi  \over 6} + {\sec ^{ - 1}}\left( {{{2(\sqrt 3  + 1)} \over {\sqrt 3 (\sqrt 3  + 1)}}} \right)$$

$$ = {\pi  \over 6} + {\sec ^{ - 1}}\left( {{2 \over {\sqrt 3 }}} \right)$$

$$ = {\pi  \over 6} + {\cos ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right)$$

$$ = {\pi  \over 6} + {\pi  \over 6}$$

$$ = {\pi  \over 3}$$

Solutions · 1