If you know that tan x = (m/n), you can find sin x.
Go about it this way: tan x = (m/n)
=> sin x/ cos x = m/n
now change cos x to sin x, use (cos x)^2 = 1 - (sin x)^2
=> sin x/ sqrt( 1- (sin x)^2) = m/n
=> (sin x)^2 / (1 - (sin x)^2) = m^2/n^2
=> (1 - (sin x)^2) / (sin x)^2 = n^2/m^2
=> 1/ (sin x)^2 - 1 = n^2/m^2
=> 1/ (sin x)^2 = n^2/m^2 +1 = (n^2 +m^2)/m^2
=> (sin x)^2 = m^2/ (n^2 + m^2)
=> sin x = sqrt [ m^2 / (n^2 + m^2)]
So sin x = m / sqrt (n^2 + m^2)
Yes, it is possible. Do not be impressed by the parametric values of the numerator and denominator of tangent function.
We know that the tangent, in a right angle triangle, is the ratio between the opposite cathetus and the joined cathetus.
In this case, is given the tan x=m/n, so we'll conclude that one cathetus is m and the other one is n.
Also, in a right angle triangle,
sin x = opposite cathetus/hypotenuse=m/hypotenuse
But in a right angle triangle, by applying Pythagorean theorem:
(hypotenuse)^2=(cathetus)^2 + (cathetus)^2
(hypotenuse)^2 = m^2 + n^2
We'll substitute the formula above
sin x =p/hypotenuse
sin x = p/sqrt (m^2 + n^2)