# Verify trigonometric equation $\frac{(\sec{A}-\csc{A})}{(\sec A+\csc A)}=\frac{(\tan A-1)}{(\tan A+1)}$

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

$$\frac{(\sin{A}-\cos{A})}{(\sin{A} \cos{A})}\div\frac{(\sin{A}+\cos{A})}{(\sin{A}\cos{A})}$$

-
I edited your question to include LaTeX formatting, and in your statement of the identity I changed "Tan+1" to "$\tan A+1$". Please Verify that this is correct. Unfortunately, for the life of me I was unable to understand what you simplified "it" too (or what "it" is). – Alex Becker Jul 13 '12 at 21:04
As a first step in situations like this, try writing everything in terms of $\sin A$ and $\cos A$, then simplify the fractions and try to use the fact that $\sin^2 A +\cos^2 A = 1$. – Old John Jul 13 '12 at 21:05
you edit it correctly – Fernando Martinez Jul 13 '12 at 21:07
@AlexBecker: I think the "it" in the question text indicates the left-hand side. So, Papi, that's a good start; notice that your simplified quotient simplifies further to $(\sin A - \cos A)/(\sin A + \cos A)$. Can you get the right-hand side to the same place? – Blue Jul 13 '12 at 21:41

Hint: Start by multiplying top and bottom on the left by $\sin A$.
@Papigrande: When you multiply $\csc A$ by $\sin A$, you will get $1$. When you multiply $\sec A$ by $\sin A$, then since $\sec A=\frac{1}{\cos A}$, you will get $\frac{\sin A}{\cos A}$, which is $\tan A$. – André Nicolas Jul 13 '12 at 21:44
$$\frac{\sec x- \csc x}{\sec x + \csc x}$$ $$=\frac{\frac{1}{\cos x} - \frac{1}{\sin x}}{\frac{1}{\cos x} + \frac{1}{\sin x}}$$ $$= \frac{\frac{\sin x}{\cos x} - 1}{\frac{\sin x}{\cos x} + 1}$$ $$= \frac{\tan x - 1}{\tan x +1}.$$