I don't really see a way to simplify this any further..
My approach to solving this is as below. I have avoided most of the algebraic manipulations in order to keep this post short and also because I feel you can do them yourself: I have concentrated on the significant steps in my method.
First, find tan(22½°) using tanA = 2.tan(A/2)/[1 - tan²(A/2)] with A = 45° and remembering that tan(45°) = 1.
We get: tan(45°) = (2.tan 22½)/(1-tan²22½) = 1
That produces a quadratic equation in tan(22½), the solutions of which are tan(22½) = -1±√2.
We also have: tan(A + B) = (tanA + tanB) / (1 - tanA.tanB).
Let A = 22½; B = 7½ and recall that tan(30°) = 1/√3.
Substitute those values for A and B in the equation immediately above, plus the calculated value for tan(22½) as just found.
The resultant algebra gives us tan(7½) = [√3.(√2 - 1) - 1] / [1 - √2 - √3].
Using a calculator, the RHS evaluates to ± 0.1317; the LHS to ±0.1318. Hence the final result is justified numerically.
The final problem is transforming [√3.(√2 - 1) - 1] / [1 - √2 - √3] to the format required by the original question. At this point, we are well away from the trig. and algebra: just a question of manipulation of surds.
For that bit, here, regretfully, I'm stuck!
Al.