Question

Hey ! do anyone know how to solve this ?

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Answer 1

`27 \csc^6\alpha + 8 \sec^6\alpha = 250`

Explanation:

`10 \sin^4 \alpha + 15\cos^4 \alpha =6, \text{ find the value of } 27 \csc^6\alpha + 8 \sec^6\alpha`

`27 \csc^6\alpha + 8 \sec^6\alpha = (27)/( \sin^6 \alpha) + (8)/( \cos^6 \alpha)`

Again, considering the identity

`\boxed{ \sin ^2 x+\cos^2 x = 1}`

then, we have
`\sin ^2 x+\cos^2 x = 1 \iff \cos^2 x = 1 -\sin ^2 x`, thus

`10 \sin^4 \alpha + 15\cos^4 \alpha = 10 \sin^4 \alpha + 15(1-\sin^2 \alpha )^2`

Considering
`x=\sin^2 \alpha`

`10 \sin^4 \alpha + 15(1-\sin^2 \alpha )^2 = 10 x^2 + 15(1-x )^2`

then,

`10 x^2 + 15(1-x )^2 = 10x^2+15(1-2x+x^2) = 25x^2-30x+15`

`25x^2-30x+15=6 \iff 25x^2-30x+9=0`

Solving using the quadratic equation:

`x =(-(-30)\pm √((-30)^2-4\cdot 25\cdot 9))/(2\cdot 25) = (30\pm √(0))/(50)`

Both roots are equal, therefore

`x=(3)/(5)\implies \sin^2\alpha =(3)/(5)`

Once we know the value of
`\sin^2\alpha`, we can find
`\cos^2\alpha`

`\sin^2 \alpha = (3)/(5) \implies (3)/(5)+\cos^2 \alpha = 1 \iff \cos^2 \alpha = (2)/(5)`

Therefore,

`\sin^2\alpha =(3)/(5) \implies \sin^6\alpha =(27)/(125)`

`\sin^2\alpha =(2)/(5) \implies \sin^6\alpha =(8)/(125)`

Finally,

`(27)/((27)/(125)) + (8)/( (8)/(125)) = 125 +125 = 250`