Question:
If $15 \sin ^{4} \alpha+10 \cos ^{4} \alpha=6$, for some $\alpha \in \mathrm{R}$, then the value of $27 \sec ^{6} \alpha+8 \operatorname{cosec}^{6} \alpha$ is equal to:
Correct Option: , 4
Solution:
$15 \sin ^{4} \alpha+10 \cos ^{4} \alpha=6$
$15 \sin ^{4} \alpha+10 \cos ^{4} \alpha=6\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}$
$\left(3 \sin ^{2} \alpha-2 \cos ^{2} \alpha\right)^{2}=0$
$\tan ^{2} \alpha=\frac{2}{3} \cdot \cot ^{2} \alpha=\frac{3}{2}$
$\Rightarrow 27 \sec ^{6} \alpha+8 \operatorname{cosec}^{6} \alpha$
$=27\left(\sec ^{6} \alpha\right)^{3}+8\left(\operatorname{cosec}^{6} \alpha\right)^{3}$
$=27\left(1+\tan ^{2} \alpha\right) 3+8\left(1+\cot ^{2} \alpha\right)^{3}$
$=250$