JEE Main 2025 Apr 4 Shift 1, Mathematics Q20: Trigonometric Ratios And Identities

Let $t={\sin }^2\theta$, so ${\cos }^2\theta =1-t$. The condition becomes $10t^2+15(1-t{)}^2=6$.

Expanding: $10t^2+15-30t+15t^2=6\Rightarrow 25t^2-30t+9=0\Rightarrow (5t-3{)}^2=0$, giving ${\sin }^2\theta =\frac{3}{5}$ and ${\cos }^2\theta =\frac{2}{5}$.

Hence ${\sin }^6\theta =\frac{27}{125}$ and ${\cos }^6\theta =\frac{8}{125}$.

Numerator: $\frac{27}{{\sin }^6\theta }+\frac{8}{{\cos }^6\theta }=\frac{27\cdot 125}{27}+\frac{8\cdot 125}{8}=125+125=250$.

Denominator: $\frac{16}{{\cos }^8\theta }=16\cdot {\left(\frac{5}{2}\right)}^4=16\cdot \frac{625}{16}=625$.

Ratio $=\frac{250}{625}=\frac{2}{5}$.