Answer:
[tex]27 \csc^6\alpha + 8 \sec^6\alpha = 250[/tex]
Step-by-step explanation:
[tex]10 \sin^4 \alpha + 15\cos^4 \alpha =6, \text{ find the value of } 27 \csc^6\alpha + 8 \sec^6\alpha[/tex]
[tex]27 \csc^6\alpha + 8 \sec^6\alpha = \dfrac{27}{ \sin^6 \alpha} + \dfrac{8}{ \cos^6 \alpha}[/tex]
Again, considering the identity
[tex]\boxed{ \sin ^2 x+\cos^2 x = 1}[/tex]
then, we have [tex]\sin ^2 x+\cos^2 x = 1 \iff \cos^2 x = 1 -\sin ^2 x[/tex], thus
[tex]10 \sin^4 \alpha + 15\cos^4 \alpha = 10 \sin^4 \alpha + 15(1-\sin^2 \alpha )^2[/tex]
Considering [tex]x=\sin^2 \alpha[/tex]
[tex]10 \sin^4 \alpha + 15(1-\sin^2 \alpha )^2 = 10 x^2 + 15(1-x )^2[/tex]
then,
[tex]10 x^2 + 15(1-x )^2 = 10x^2+15(1-2x+x^2) = 25x^2-30x+15[/tex]
[tex]25x^2-30x+15=6 \iff 25x^2-30x+9=0[/tex]
Solving using the quadratic equation:
[tex]x =\dfrac{-(-30)\pm \sqrt{(-30)^2-4\cdot 25\cdot 9}}{2\cdot 25} = \dfrac{30\pm \sqrt{0}}{50}[/tex]
Both roots are equal, therefore
[tex]x=\dfrac{3}{5}\implies \sin^2\alpha =\dfrac{3}{5}[/tex]
Once we know the value of [tex]\sin^2\alpha[/tex], we can find [tex]\cos^2\alpha[/tex]
[tex]\sin^2 \alpha = \dfrac{3}{5} \implies \dfrac{3}{5}+\cos^2 \alpha = 1 \iff \cos^2 \alpha = \dfrac{2}{5}[/tex]
Therefore,
[tex]\sin^2\alpha =\dfrac{3}{5} \implies \sin^6\alpha =\dfrac{27}{125}[/tex]
[tex]\sin^2\alpha =\dfrac{2}{5} \implies \sin^6\alpha =\dfrac{8}{125}[/tex]
Finally,
[tex]\dfrac{27}{\frac{27}{125}} + \dfrac{8}{ \frac{8}{125}} = 125 +125 = 250[/tex]
