\(\sin ^{6} \theta+\cos ^{6} \theta=\frac{1}{4} ;

\frac{1}{\sin ^{6} \theta}+\frac{1}{\cos ^{6} \theta}=\) ?

We have \(\sin ^{6} \theta+\cos ^{6} \theta=\frac{1}{4}\).

\[

\begin{aligned}

&\Rightarrow\left(\sin ^{2} \theta\right)^{3}+\left(\cos ^{2} \theta\right)^{3}=\frac{1}{4} \\

&\Rightarrow\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-3 \sin ^{2} \theta \cos ^{2} \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\frac{1}{4} \\

&\Rightarrow 1-3 \sin ^{2} \theta \cos ^{2} \theta=\frac{1}{4} \Rightarrow 3 \sin ^{2} \theta \cos ^{2} \theta=\frac{3}{4} \\

&\therefore \sin ^{2} \theta \cos ^{2} \theta=\frac{1}{4} \Rightarrow \sin ^{6} \theta \cos ^{6} \theta=\frac{1}{64} . \\

&\text { Thus, } \frac{1}{\sin ^{6} \theta}+\frac{1}{\cos ^{6} \theta}=\frac{\sin ^{6} \theta+\cos ^{6} \theta}{\sin ^{6} \theta \cos ^{6} \theta}=\frac{1}{\frac{1}{1}}=16 .

\end{aligned}

\]