# 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.

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?