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

$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}$
 A: $$
x=\sec A + \tan A= \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}
$$
so we have:
$$
x+\frac{1}{x}= \frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=\frac{1+\sin^2 A+2\sin A+\cos^2 A}{(\cos A)(1+ \sin A)}=
$$
$$
=\frac{2(1+\sin A)}{(\cos A)(1+ \sin A)}=\frac{2}{\cos A}
$$
A: Because $$x+\frac{1}{x}=\frac{1+\sin{A}}{\cos{A}}+\frac{\cos{A}}{1+\sin{A}}=$$
$$=\frac{(1+\sin{A})^2}{\cos{A}(1+\sin{A})}+\frac{\cos^2{A}}{\cos{A}(1+\sin{A})}=\frac{(1+\sin{A})^2+\cos^2A}{\cos{A}(1+\sin{A})}=$$
$$=\frac{1+2\sin{A}+\sin^2A+\cos^2A}{\cos{A}(1+\sin{A})}=\frac{1+2\sin{A}+1}{\cos{A}(1+\sin{A})}=$$
$$=\frac{2+2\sin{A}}{\cos{A}(1+\sin{A})}=\frac{2}{\cos{A}}=2\sec{A}.$$
A: x=secA + tanA
1/x= 1/(secA + tanA).                                               = (secA - tanA)/(secA-tanA)(secA+tanA)
=(secA - tanA)/{(secA)^2 - (tanA)^2}
= secA - tanA
Therefore
x+ 1/x = (secA + tanA)+(secA - tanA)
            = 2SecA
A: Enough to show:
$$\cos{A}\left( x + \frac{1}{x}\right) = 2$$
$$\cos{A} \cdot x + \frac{\cos{A}}{x} = 1 +\sin{A} + \frac{\cos^2{A}}{1 +\sin{A}} = 1 +\sin{A} + \frac{1-\sin^2{A}}{1 +\sin{A}} = 2 $$
A: we get $$x+\frac{1}{x}={\frac { \left( \sec \left( A \right)  \right) ^{2}+2\,\sec \left( A
 \right) \tan \left( A \right) + \left( \tan \left( A \right) 
 \right) ^{2}+1}{\sec \left( A \right) +\tan \left( A \right) }}
$$
can you simplify this term?
the numerator simplifies to $$-\frac{2}{\sin(A)-1}$$
and $$\sec(A)+\tan(A)=\frac{\sin(A)+1}{\cos(A)}$$
