Why cant I prove this trigonometric equation straight down? The question is as follows:
Given that $\tan^2a - 2 \tan^2b = 1$. Show that $\cos2a + \sin^2b = 0$.
After a few attempts, I successfully came up with a solution as follows:
$$\tan^2a - 2 \tan^2b = 1$$
$$(\sec^2a - 1) - 2(\sec^2b - 1) = 1$$
$$\sec^2a - 2sec^2b = 0$$
$$\frac{1}{\cos^2a} - \frac{2}{\cos^2b} = 0$$
$$\cos^2b - 2\cos^2a= 0$$
$$(1-\sin^2b) - (1 - \sin^2a) - \cos^2a = 0$$
$$\sin^2a - \cos^2a + \sin^2b = 0$$
$$\cos2a + \sin^2b = 0$$
And the proof is finished. However, before I found this solution, I came up with something like this:
$$\tan^2a - 2 \tan^2b = 1$$
$$\frac{\sin^2a}{\cos^2a} - 2\frac{\sin^2b}{\cos^2b} = 1$$
$$\sin^2a\cos^2b - 2\sin^2b\cos^2a - \cos^2a\cos^2b = 0$$
I wanted to extract the $\cos2a $ here.  So I do the following
$$(\sin^2a - \cos^2a) \cos^2b - 2\sin^2b\cos^2a = 0$$
$$\cos2a + \frac{2\sin^2b\cos^2a}{\cos^2b} = 0$$
It suffices to show that
$$\frac{2\sin^2b\cos^2a}{\cos^2b} = \sin^2b$$
which is equalvalent as saying 
$$\frac{\cos^2a}{\cos^2b} = \frac{1}{2}$$
Therefore, I looked at the original equation and proved that it is true like this:
$$\tan^2a - 2 \tan^2b = 1$$
$$(\sec^2a - 1) - 2(\sec^2b - 1) = 1$$
$$\sec^2a = 2sec^2b$$
$$\frac{\cos^2a}{\cos^2b} = \frac{1}{2}$$
My confusion is the reason why I need to look at the orginal equation again to finish the proof. $\tan^2a - 2 \tan^2b = 1$ is identical to  $\cos2a + \sin^2b = 0$. And I didn't lose information during my proof. Shouldn't I be able to finish the proof directly like my first proof?
It may sounds like a strange question. But I hope that someone can explain it to me because it sometimes happens and is confusing to me. 

Starting from this line continue as follows (instead of dividing by $\cos ^2b$, express it as $1-\sin^2b$):
$$(\sin^2a - \cos^2a) \cos^2b - 2\sin^2b\cos^2a = 0 \\
(\sin^2a - \cos^2a) (1-\sin^2b) - 2\sin^2b\cos^2a = 0\\
-\cos 2a-\sin^2a\sin^2b+\sin^2b\cos^2a-2\sin^2b\cos^2a=0 \\
-\cos2a-\sin^2a\sin^2b-\sin^2b\cos^2a=0\\
\cos2a+\sin^2b=0.$$
