Circular functions question help?

[Yoda]

Need help with the following questions thanks

a) Use a compound angle formula to show that tan^-1(3) - tan^-1(1/2) = pi/4

b) Hence show that tan^-1(x) - tan^-1(x-1/x+1) = pi/4 , x>-1.

[RKTR]

a) let tan^-1(3)=x, tan^-1(1/2)=y
          tan(x)=3, tan(y)=1/2

LHS:  tan^-1(tan[tan^-1(3)-tan^-1(1/2)]
        = tan^-1 [tan(x-y)]
         =tan^-1 [ tanx -tany /(1+tanx tany)
         =tan^-1 ( 3-1/2 )/(1+ 3(1/2)
         =tan^-1 [(5/2) / (5/2)]
         =tan^-1(1)
         =pi/4
         =RHS (Shown)
 
b) let tan^-1(x)=a ,tan(a)=x
         tan^-1(x-1/x+1)=b,tan(b)=x-1/x+1

LHS: tan^-1(x)-tan^-1(x-1/x+1)
       =tan^-1(tan[tan^-1(x)-tan^-1(x-1/x+1)
       =tan^-1(tan(a-b)
        =tan^-1[(tan a -tanb)/(1+tana tan b)]
        =tan^-1[x -( x-1)/(x+1)]/[1+ (x)(x-1)/(x+1)]
        =tan^-1{[(x^2+x-x+1)/(x+1)]/ [(x+1+x^2-x)/(x+1)]
        =tan^-1 (x^2+1)/(x^2+1)
        =tan^-1(1)
        =pi/4
        =RHS (Shown)