write the value of tan-1[2 sin {2 cos-1(root 3/ 2)}]

Answer  = tan^-1[2 sin {2 cos-1(3^1/2/ 2)}] 

= tan^-1[2 sin {2 (pi/6)}]  

     {as we know cos pi/6 =   3 ^1/2/2  so cos ^-1 (3 ^1/2/2 ) = pi/6  }

=  tan^-1[2 sin (pi/3)]

= tan^-1 [ 2 (3 ^1/2/2 ) ]                       {sin pi / 3 = 3 ^1/2/2 }

= tan^-1 [ 3 ^1/2]

= pi/3  Answer