the number of real solutions of the eq         

take the  2 to RHS

so we will have  sec ^-1(x) = pie/2 - 2 tan ^-1(x)

so   x = sec ( pie/2 -  2 tan ^-1(x) )

ie x = cosec ( 2 tan ^-1(x) )

now put x= tan    --------------  1

so we get    tan = cosec (2)

ie sin / cos = 1/ 2 sin cos

ie 2 sin ² = 1

or sin = +- 1/ root 2

putting it in the 1st equation we get

x =1

oops,

Anchit's solution is correct