JEE Main 2025 — 4 April, Morning Shift — Question 29

$10 \sin ^{4} \theta+15 \cos ^{4} \theta=6$

$\Rightarrow 10 \sin ^{4} \theta+10 \cos ^{4} \theta+5 \cos ^{4} \theta=6$

$\Rightarrow 10\left[\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \cos ^{2} \theta\right]+5 \cos ^{4} \theta=6$

$\Rightarrow 10-20\left(1-\cos ^{2} \theta\right) \cos ^{2} \theta+5 \cos ^{4} \theta=6$ Let

$\cos ^{2} \theta=x$

$$10-20\left(x-x^{2}\right)+5 x^{2}=6$$ $$\begin{aligned} & \Rightarrow 25 x^{2}-20 x+4=0 \\& (5 x-2)^{2}=0 \Rightarrow x=\frac{2}{5} \\& \Rightarrow \quad \cos ^{2} \theta=\frac{2}{5} \Rightarrow \sin ^{2} \theta=\frac{3}{5} \\& \sec ^{2} \theta=\frac{5}{2}, \operatorname{cosec}^{2} \theta=\frac{5}{3} \\& \frac{27 \operatorname{cosec}^{6} \theta+8 \sec ^{6} \theta}{16 \sec ^{8} \theta}=\frac{27\left(\frac{5}{3}\right)^{3}+8\left(\frac{5}{2}\right)^{3}}{16\left(\frac{5}{2}\right)^{4}} \\& =\frac{5^{3}+5^{3}}{5^{4}}=\frac{2.5^{3}}{5^{4}}=\frac{2}{5} \end{aligned}$$