2
$\begingroup$

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

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

  • 0
    @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?2012-07-13

2 Answers 2

2

Hint: Start by multiplying top and bottom on the left by $\sin A$.

  • 1
    @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 $\tan A$.2012-07-13
3

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