Trigonometric equation: $2 \cot^2 \alpha - 2 \sec^2 \alpha = 1$ Can someone help me solving this equation:

$$
2 \cot^2 \alpha - 2 \sec^2 \alpha = 1
$$

I have to solve this for $\alpha$. Is there any relation I'm missing? I guess I'm going to have to bring both to the same trigonometric number?
Hint

$$\sec^2\alpha=1+\tan^2\alpha$$

so

$$\frac{1}{\tan^2\alpha}-1-\tan^2\alpha=\frac{1}{2}$$

Now use $u=\tan^2\alpha$

$$\frac{1}{u}-u-\frac{3}{2}=0 \rightarrow 2u^2+3u-2=0$$

What give us $u=-2$ or $u=1/2$.

Can you finish?
$$
\sec^2(\alpha)=1+\tan^2(\alpha)
$$
And
$$
\cot^2(\alpha)=\frac{1}{\tan^2(\alpha)}
$$
Simplifying we get a quadratic equation which can solved to get the value of $\tan^2(\alpha)$