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.
Best Answer
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 $\sin^2u+\cos^2 u=1$ $$\sqrt{\frac{\sin^2 u\cos^2 t+\cos^2u\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}=\sqrt{\frac{(1-\cos^2 u)\cos^2 t+(1-\sin^2u)\sin^2t}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{(\cos^2 t+\sin^2 t)-(\cos^2 u\cos^2 t+\sin^2u\sin^2t)}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}}\\=\sqrt{\frac{1}{\cos^2 u\cos^2 t+\sin^2u\sin^2t}-1}$$