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I am looking to simplify these term [ I forgot the 3 :( ]
$$\sqrt\frac{\left(a^2\cos^2t+b^2\sin^2t\right)^3}{\left(b^2\cos^2t+a^2\sin^2t\right)^3}$$
where $a$ and $b$ are two non-negative reals.
(This is not homework. I am just trying to make my expression easy, but I didn't find a way.)
Thanks for your help.
The form is $$\sqrt{\frac{f^3}{g^3}}$$ which we can write as $$\left(\frac{f}{g}\right)^{3/2}$$ so let's just worry about that inner quotient, $f/g$.
The quotient is definitely not constant; different $t$ values give different results: $$t = 0 \;\to\; \frac{a^2}{b^2} \qquad\qquad t= \frac{\pi}{2}\;\to\;\frac{b^2}{a^2}$$
Having to set aside this dream case, we're left to consider a few alternatives and decide which might be consider least-bad.
Leave it as-is. $$\frac{a^2\cos^2 t + b^2 \sin^2 t}{a^2 \sin^2 t+b^2\cos^2 t} \tag{1}$$ That expression isn't terribly complicated.
Divide-through by $\cos^2t$ in the numerator and denominator, to get $$\frac{a^2+b^2\tan^2t}{b^2+a^2\tan^2t} \tag{2}$$ This may not be appreciably better, though ... and it introduces unnecessary concern about $t=\pi/2$.
@Dr.SonnhardGraubner's answer invokes the double-angle formula to get something I'll write as $$\frac{\left(a^2+b^2\right)+\left(a^2-b^2\right)\cos 2t}{\left(a^2+b^2\right)-\left(a^2-b^2\right)\cos 2t} \tag{3}$$ which is "simpler" in that the degree of the trig functions is lower, at the cose of adding some complexity to the coefficients.
@Andrei's suggestion to trade $a$ and $b$ for trig functions is a good one, although I'd choose to swap the sine and cosine assignment to write $a = \sqrt{a^2+b^2} \cos u$ (and $b = \sqrt{a^2+b^2} \sin u$). I'd also use the double-angle formulas to simplify the resulting quotient to $$\frac{1 + \cos 2t \cos 2u}{1 - \cos 2t \cos 2u} \tag{4}$$ (I'd also probably choose to associate $a$ with $\cos u$ (and $b$ with $\sin u$), which changes the above slightly.) Note that $(3)$ seems to be crying-out for us to make such a substitution.
Since the quotient appears in the context of ellipses, we could use $a^2-b^2=c^2$ (where $c$ is the center-to-focus distance) and $c = ea$ (where $e$ is the eccentricity) to write $$\frac{ 1 -e^2 \sin^2 t}{1 -e^2 \cos^2 t} \tag{5}$$ I like this one, personally.
One can also combine re-writing in terms of $e$ with re-writing in terms of $\cos 2t$ (left as an exercise to the reader), but this just seems to re-complicate things.
One can imagine other variations, too. How useful any one version is depends upon how it's intended to be used.
Divide both the numerator and denominator by $a^2+b^2$, then use $\frac{a^2}{a^2+b^2}=\sin^2 u$. Your expression will become $$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}$$ Now use $\sin^2u+\cos^2 u=1$ $$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}=\sqrt{\frac{(1-\cos^2 u)\cos^2 t+(1-\sin^2u)\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{(\cos^2 t+\sin^2 t)-(\cos^2 u\cos^2 t+\sin^2u\sin^2t)}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{1}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}-1}$$
For the radicand i have got $$\frac{a^2 (-\cos (2 t))-a^2+b^2 \cos (2 t)-b^2}{a^2 \cos (2 t)-a^2-b^2 \cos (2 t)-b^2}$$
