If 10 sin4⁡α+15cos4⁡α=6 then the value of 27cosec6⁡α+8sec6⁡α is

Solution:

We have,

$\begin{array}{l} 10 \sin^{4} α + 15 \cos^{4} α = 6 \\ \Rightarrow 10 \sin^{4} α + 15 \cos^{4} α = 6 {(\sin^{2} α + \cos^{2} α)}^{2} \\ \Rightarrow 10 \tan^{4} α + 15 = 6 {(\tan^{2} α + 1)}^{2} \\ \Rightarrow {(2 \tan^{2} α - 3)}^{2} = 0 \Rightarrow \tan^{2} α = \frac{3}{2} \end{array}$

$\begin{array}{l} ∴ 27 {cosec}^{6} α + 8 \sec^{6} α \\ = 27 {(1 + \cot^{2} α)}^{3} + 8 {(1 + \tan^{2} α)}^{3} \\ = 27 {(1 + \frac{2}{3})}^{3} + 8 {(1 + \frac{3}{2})}^{3} = 250 \end{array}$

If $10 \sin^{4} α + 15 \cos^{4} α = 6$ then the value of $27 {cosec}^{6} α + 8 \sec^{6} α$ is

Solution:

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