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

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

$$
x=\sec A + \tan A= \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}
$$
so we have:
$$
x+\frac{1}{x}= \frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=\frac{1+\sin^2 A+2\sin A+\cos^2 A}{(\cos A)(1+ \sin A)}=
$$
$$
=\frac{2(1+\sin A)}{(\cos A)(1+ \sin A)}=\frac{2}{\cos A}
$$
