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