<s>[INST] 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? 
 [/INST] Following your way:
$$10 \sin^4\theta + 15 \cos^4 \theta=6(\sin^2\theta + \cos \theta^2)^2$$
$$\iff4(\sin^2\theta)^2-12\sin^2\theta\cos^2\theta+9(\cos^2\theta)^2=0$$
Dividing either sides by $\cos^4 \theta,$
$$4(\tan^2\theta)^2-12\tan^2\theta+9=0\iff(2\tan^2\theta-3)^2=0$$
We can find $\sec^2\theta=1+\tan^2\theta, \cot^2\theta=\dfrac1{\tan^2\theta},\csc^2\theta=1+\cot^2\theta$
Hope you can take it home from here
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