Secant and Tangent identity i've been stuck on this question too long
$x = \sec A + \tan A$
show $x + \frac{1}{x} = 2\cdot \sec A$
I've been using $\tan^2 \theta + 1 = \sec^2 \theta$
and $\tan\theta = \frac{\sin \theta}{\cos\theta}$
help would be much appreciated

$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}$

Enough to show:
$$\cos{A}\left( x + \frac{1}{x}\right) = 2$$
$$\cos{A} \cdot x + \frac{\cos{A}}{x} = 1 +\sin{A} + \frac{\cos^2{A}}{1 +\sin{A}} = 1 +\sin{A} + \frac{1-\sin^2{A}}{1 +\sin{A}} = 2 $$
