<s>[INST] Find the value of $27\csc^2\theta+8\sec^2\theta$ $10\sin^4\theta+15\cos^4\theta=6$, then find the value of  $27\csc^2\theta+8\sec^2\theta$
I don't know how to do it  have just tried by converting sin and cos into csc and sec. But can't get the answer.
 [/INST] Since $\sin^2\theta+\cos^2\theta=1$,
$$10\sin^4\theta+15\cos^4\theta=6(\sin^2\theta+\cos^2\theta)^2.$$
It follows that
$$4\sin^4\theta-12\sin^2\theta\cos^2\theta+9\cos^4\theta=0,$$
or 
$$(2\sin^2\theta-3\cos^2\theta)^2=0.$$
Thus we have $2\sin^2\theta=3\cos^2\theta$. Can you continue from here?
</s>