JEE Main 2021 — Mathematics Trigonometry

$$15{\mathrm{sin}}^{4}\alpha +10{\mathrm{cos}}^{4}\alpha =6$$15sin4α+10cos4α=6

$$15{\mathrm{sin}}^{4}\alpha +10{\mathrm{cos}}^{4}\alpha =6{({\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha )}^{2}$$15sin4α+10cos4α=6(sin2α+cos2α)2

$${(3{\mathrm{sin}}^{2}\alpha -2{\mathrm{cos}}^{2}\alpha )}^{2}=0$$(3sin2α−2cos2α)2=0

$$\therefore 3{\mathrm{sin}}^{2}\alpha -2{\mathrm{cos}}^{2}\alpha =0$$∴3sin2α−2cos2α=0

{\mathrm{tan}}^{2}\alpha =\frac{2}{3}&{\mathrm{cot}}^{2}\alpha =\frac{3}{2}

$$\Rightarrow 27{\mathrm{sec}}^{6}\alpha +8{cosec}^{6}\alpha$$⇒27sec6α+8cosec6α

$$=27{({\mathrm{sec}}^{2}\alpha )}^{3}+8{({cosec}^{2}\alpha )}^{3}$$=27(sec2α)3+8(cosec2α)3

$$=27{(1+{\mathrm{tan}}^{2}\alpha )}^{3}+8{(1+{\mathrm{cot}}^{2}\alpha )}^{3}$$=27(1+tan2α)3+8(1+cot2α)3

$$=27{(1+\frac{2}{3})}^{3}+8{(1+\frac{3}{2})}^{3}$$=27(1+32​)3+8(1+23​)3

$$=250$$=250