<s>[INST] 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

$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}$
 [/INST] Because $$x+\frac{1}{x}=\frac{1+\sin{A}}{\cos{A}}+\frac{\cos{A}}{1+\sin{A}}=$$
$$=\frac{(1+\sin{A})^2}{\cos{A}(1+\sin{A})}+\frac{\cos^2{A}}{\cos{A}(1+\sin{A})}=\frac{(1+\sin{A})^2+\cos^2A}{\cos{A}(1+\sin{A})}=$$
$$=\frac{1+2\sin{A}+\sin^2A+\cos^2A}{\cos{A}(1+\sin{A})}=\frac{1+2\sin{A}+1}{\cos{A}(1+\sin{A})}=$$
$$=\frac{2+2\sin{A}}{\cos{A}(1+\sin{A})}=\frac{2}{\cos{A}}=2\sec{A}.$$
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