Any shorter way to solve trigonometric problem? If $10 \sin^4\theta + 15 \cos^4 \theta=6$, then find value of $27 \csc^2 \theta + 8\sec^2 \theta$
I know the normal method o solve this problem in which we need to multiply L.H.S. of $10 \sin^4\theta + 15 \cos^4 \theta=6$ by $(\sin^2\theta + \cos \theta^2)^2$ and then simplifying it. It is kind of simplification in which we can commit silly mistakes.
So, is there any shorter way to solve this trigonometric problem? 

How can $20\sin^2\theta + 15\cos^2\theta = 6$? $20\sin^2\theta + 15 \cos^2\theta = 15 + 5\sin^2\theta$, implying that $5 \sin^2\theta = -9$, but $\sin^2\theta$ should be nonnegative.
