Simplify absolute value in math

Question

I currently have a great expression with many form members

Abs[-2 b + 2 d1 m + l Tan[\[Theta]]]

I know from the geometry of my problem that

-2 b + 2 d1 m + l Tan[\[Theta]] > 0

However, when I try to simplify my expression,

Simplify[Abs[-2 b + 2 d1 m + l Tan[\[Theta]]], -2 b + 2 d1 m + l Tan[\[Theta]] > 0]

I'll just be back

Abs[-2 b + 2 d1 m + l Tan[\[Theta]]]

How can I get Mathematica to simplify an unnecessary absolute value?

EDIT 1

The full expression I'm trying to simplify

-(1/(2 (m - Tan[\[Theta]])))
 Sqrt[1 + m^2] (B2 Sqrt[(-2 b + 2 d1 m + l Tan[\[Theta]])^2] + 
    B4 Sqrt[(-2 b + 2 d2 m + l Tan[\[Theta]])^2] + 
    B5 Sqrt[(2 b + 2 d3 m + l Tan[\[Theta]])^2] + 
    B7 Sqrt[(2 b + 2 d4 m + l Tan[\[Theta]])^2] + 
    B1 Sqrt[(2 b - 2 (d1 + l) m + l Tan[\[Theta]])^2] + 
    B3 Sqrt[(2 b - 2 (d2 + l) m + l Tan[\[Theta]])^2] + 
    B6 Sqrt[(-2 (b + (d3 + l) m) + l Tan[\[Theta]])^2] + 
    B8 Sqrt[(-2 (b + (d4 + l) m) + l Tan[\[Theta]])^2])

The terms squared under each of the radicals are known to be a positive real number.

Answer 1

Since all terms are known as real and positive, squaring and taking the square root will give you only the same number. Therefore, you can do something like

expr /. Sqrt[(x___)^2] :> x

expr - .

Answer 2

:

1)

Simplify[Abs[-2 b + 2 d1 m + l Tan[\[Theta]]], 
 0 < \[Theta] < \[Pi]/2 && l > 0 && 2 d1 m > 0 && -2 b > 0]

2)

f[e_] := 100 Count[e, _Abs, {0, Infinity}] + LeafCount[e]
Simplify[Abs[-2 b + 2 d1 m + l Tan[\[Theta]]], -2 b + 2 d1 m + 
   l Tan[\[Theta]] > 0, ComplexityFunction -> f]

Th f Abs , Times. . . ?

Answer 3

, - :

Clear[removeAbs]
removeAbs[expr_, r_] := expr /. {Sqrt[r^2] :> r, Abs[r] :> r}

, , :

In: removeAbs[Abs[x] + Abs[y], x]
Out: x + Abs[y]

, , .

Answer 4

, Abs[a]^2, , Assuming a\[Element]Reals, .

WolframMathWorld - Absolute Square ComplexExpand[Abs[a]^2, TargetFunctions -> {Conjugate}], Conjugate[Sqrt[a^2 + b^2]] ComplexExpand ( ) , .