How do i find $\tan(\theta)$ such that : $\frac{16}{\sin^6(\theta)} + \frac{81}{\cos^6(\theta)}=625$?? How do i find $\tan(\theta)$ such that :$$\frac{16}{\sin^6(\theta)} + \frac{81}{\cos^6(\theta)}=625$$?
Note : i used some trigono-form but sorry i didn't succed .
Thank you for any help.

I cannot say that this is an elegant solution but oh well,
i will use $s$ for $sin$, $c$ for $cos$ and $t$ for tan
we know that $\frac{s}{c} = t$
so $\frac{16}{c^{6}*t^{6}} + \frac{81}{c^{6}} = 625$
$16 + 81t^{6} = 625c^{6}*t^{6}$
we know that $ 1 + t^{2} = \frac{1}{c^{2}}$
or $16 + 81t^{6} = \frac{625*t^{6}}{(t^{2} +1)^{3}}$
simplifying gives
$81t^{12} + 243t^{10} + 243t^{8} + - 528t^{6} + 48t^{4} + 48t^{2} +16 = 0$
let $y = t^{2}$
$81y^{6} + 243y^{5} + 243y^{4} + - 528y^{3} + 48y^{2} + 48y^{1} +16 = 0$
using the rational root theorem, we find that $\frac{2}{3}$ works
so $\sqrt{\frac{2}{3}} = \boxed{\tan(x) = \frac{\sqrt{6}}{3}}$
