I have two trigonometric expressions. I would like to eliminate the fraction from my 2nd expression.

A = (-Cos[ψ2] Sin[η2] Sin[θ2] + Cos[η2] Sin[ψ2]) 

B = (-D + d Cos[η2] Cos[θ2] Cos[ϕ2] + 1/2 l Cos[θ2] Sin[η2] + Q2 Sin[θ2] - d Sin[θ2] Sin[ϕ2] + (Q1 Cos[θ1] Cos[θ2] Sin[η2] (Cos[η2] Cos[θ2] Cos[ϕ2] - Sin[θ2] Sin[ϕ2])) / (Cos[ψ2] Sin[η2] Sin[θ2] - Cos[η2] Sin[ψ2]))

I tried Simplify and FullSimplify to suppress the denominator but the denominator in the expression B (equal to A) is not suppressed.


Please help me to force Mathematica to make the simplifications so as to eliminate the denominator of the expression B which is equal to A ?

  • 3
    $\begingroup$ Try Simplify@Expand[A*B] $\endgroup$
    – Lukas Lang
    May 30, 2018 at 13:54
  • $\begingroup$ Or Cancel[A*B]. $\endgroup$
    – John Doty
    May 30, 2018 at 15:14

1 Answer 1


First let me explain what is going on.

 A // FullForm
Plus[Times[-1, Cos[ψ2], Sin[η2], Sin[θ2]], Times[Cos[η2], Sin[ψ2]]]

However, what you consider the denominator of the fractional part of b is

(Cos[ψ2] Sin[η2] Sin[θ2] - Cos[η2] Sin[ψ2]) // FullForm

Plus[Times[Cos[ψ2], Sin[η2], Sin[θ2]], Times[-1, Cos[η2], Sin[ψ2]]]

So the two forms don't look the same to Mathematica and don't get cancelled.

There are many ways to solve the problem. One has already been suggested in a comment. Here is another.

AA = -1 A;
Simplify[AA B]



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