If $7\sin^2A-3\cos^2A=4$, then $\tan A\neq\pm\frac{1}{\sqrt{3}}$ 
If $7\sin^2A-3\cos^2A=4$, then we have to show that $\tan A=\pm\dfrac{1}{\sqrt{3}}$

The textbook approaches this by substituting $\sin^2A=1-\cos^2A$ and that method does(not) work.
$7\sin^2A-3\cos^2A=4$ 
$\implies  7\sin^2A-3(1-\sin^2A)=4$ {contains a sign error)
$\implies \sin^2a=\dfrac{1}{4}$
$\implies \cos^2A=1-\sin^2A=\dfrac{3}{4}$
Then, $\tan^2A=\dfrac{\sin^2A}{\cos^2A}=\dfrac{1}{3}$
$\implies \tan A=\pm\dfrac{1}{\sqrt{3}}$
But what's wrong with mine?(nothing)
$7\sin^2A-3\cos^2A=4(\sin^2A+\cos^2A)$
$\implies 3\sin^2A=7\cos^2A$
$\implies \dfrac{\sin^2A}{\cos^2A}=\dfrac{7}{3}$
$\implies \tan^2A=\dfrac{7}{3}$
$\implies \tan A=\sqrt{\dfrac{7}{3}}$
EDIT:Actually,the  textbook(as usual) made a sign error in the second line.Ignore it.My solution is correct.

The book made a sign error in the second line.My method is correct.
