If $3\sec^4\theta+8=10\sec^2\theta$, find the values of $\tan\theta$ If $3\sec^4\theta+8=10\sec^2\theta$, find the values of $\tan\theta$.
$$3\sec^4\theta-10\sec^2\theta+8=0$$
$$3\sec^4\theta-6\sec^2\theta-4\sec^2\theta+8=0$$
$$(3\sec^2\theta-4)(\sec^2\theta-2)=0$$
$$\sec^2\theta=2 \text { or } \sec^2\theta=\frac{4}{3}$$
$$1+\tan^2\theta=2 \text { or } 1+\tan^2\theta=\frac{4}{3}$$
$$\tan^2\theta=1 \text { or } \tan^2\theta=\frac{1}{3}$$
$$\begin{equation}
\tan\theta=\pm1  \text { or } \tan\theta=\pm\frac{1}{\sqrt{3}}
\end{equation}$$
So $\tan\theta$ can have four values $1,-1,\dfrac{1}{\sqrt{3}},-\dfrac{1}{\sqrt{3}}$.
But answer is only $1 \text{ or } \dfrac{1}{\sqrt{3}}$.
What are the things I am missing here?

Your method is correct. Whatever source told you only the positive roots are right either made a mistake or imposed (or meant to impose) a range on $\theta$ that ensures $\tan\theta>0$, such as $\theta\in\left(0,\,\frac{\pi}{2}\right)$.
