SecA TanA
----- - ----- = 1
CosA Cot A
Please help, I don't know where to start or how to complete this.
Think of the pythagorean identity $\displaystyle tan^2(x) + 1 = sec^2(x)$
See how it can be re-arranged to look like $\displaystyle sec^2(x) - tan^2(x) = 1$
Now, how are $\displaystyle \frac{sec(x)}{cos(x)}$ and $\displaystyle sec^2(x)$ related?
How are $\displaystyle \frac{tan(x)}{cot(x)}$ and $\displaystyle tan^2(x)$ related?
Hello, hachm361!
$\displaystyle \text{Prove: }\:\frac{\sec A}{\cos A} - \frac{\tan A}{\cot A} \:=\:1$
Do you know any basic identities . . . like these?
. . $\displaystyle \cos A \,=\,\frac{1}{\sec A} \qquad \cot A \,=\,\frac{1}{\tan A} \qquad \sec^2\!A - \tan^2\!A \:=\:1$
If you don't, you need more help than we can offer.
We have: .$\displaystyle \frac{\sec A}{\cos A} - \frac{\tan A}{\cot A} \;=\;\frac{\sec A}{\frac{1}{\sec A}} - \frac{\tan A}{\frac{1}{\tan A}} \;=\;\sec^2\!A - \tan^2\!A \;=\;1$