If 48 sin4 ∝ + 80 cos4 ∝ = 30, then 54 tan6 ∝ + 125 cosec6 ∝ = ?

\(\frac{48}{30}\) sin⁴ ∝ + \(\frac{80}{30}\) cos⁴ ∝ = 1

( \(\frac{8}{5}\)sin² ∝) . sin² ∝ + ( \(\frac{8}{3}\)cos² ∝) . cos² ∝ - 1

we know 1. sin² ∝ + cos² ∝ = 1

so, \(\frac{8}{5}\)sin² ∝ = 1               cos² ∝ = \(\frac{3}{8}\) = tan² ∝ = \(\frac{5}{3}\)

sin² ∝ = \(\frac{5}{8}\)

54tan⁶ ∝+ 125 cosec⁶ ∝ = 54 × (\(\frac{125}{27}\)) + 125 (\(\frac{512}{125}\)) = 250 + 512 = 762