Given:
\[\tan A = 3 \cot A \implies \tan A = 3 \times \frac{1}{\tan A}\]
\[\tan^2 A = 3 \implies \tan A = \sqrt{3}\]
For $\tan A = \sqrt{3}$, the corresponding angle is:
\[A = 60^\circ\]
If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is: