How Would I verify the following identity.
$\frac{(\sec{A}-\csc{A})}{(\sec A+\csc A)}=\frac{(\tan A-1)}{(\tan A+1)}$
I simplified it to
$\frac{(\sin{A}-\cos{A})}{(\sin{A} \cos{A})}\div\frac{(\sin{A}+\cos{A})}{(\sin{A}\cos{A})}$
How Would I verify the following identity.
$\frac{(\sec{A}-\csc{A})}{(\sec A+\csc A)}=\frac{(\tan A-1)}{(\tan A+1)}$
I simplified it to
$\frac{(\sin{A}-\cos{A})}{(\sin{A} \cos{A})}\div\frac{(\sin{A}+\cos{A})}{(\sin{A}\cos{A})}$
Hint: Start by multiplying top and bottom on the left by $\sin A$.
$\frac{\sec x- \csc x}{\sec x + \csc x}$ $=\frac{\frac{1}{\cos x} - \frac{1}{\sin x}}{\frac{1}{\cos x} + \frac{1}{\sin x}}$ $ = \frac{\frac{\sin x}{\cos x} - 1}{\frac{\sin x}{\cos x} + 1}$ $ = \frac{\tan x - 1}{\tan x +1}.$