# Thread: Trig Identities in longer way

1. ## Trig Identities in longer way

I have this trig problem:

$\frac {secx - 1}{Tan x } = \frac {Tan x}{Sec x - 1}$

now what we did in class is cross multuplying both of them and we get the answer.

I want to know how to get the answer in long process. thank you!

2. Hello Anemori
Originally Posted by Anemori
I have this trig problem:

$\frac {secx - 1}{Tan x } = \frac {Tan x}{Sec x - 1}$

now what we did in class is cross multuplying both of them and we get the answer.

I want to know how to get the answer in long process. thank you!
I think you have a sign wrong. One of the minus signs should be a plus sign. So perhaps you must mean: prove the identity:
$\frac {\sec x - 1}{\tan x } = \frac {\tan x}{\sec x + 1}$
Start with the LHS, and multiply top-and-bottom of the fraction by $(\sec x + 1)$
$\frac {\sec x - 1}{\tan x } =\frac {(\sec x - 1)\color{red}(\sec x + 1)}{\tan x\color{red} (\sec x + 1)}$
$=\frac{\sec^2x-1}{\tan x (\sec x + 1)}$

$=\frac{\tan^2x}{\tan x (\sec x + 1)}$, using the identity $\sec^2x -1 = \tan^2x$

$=\frac{\tan x}{\sec x + 1}$

On the other hand, if it's the other minus sign that's wrong, again you'll start with the LHS, but this time multiply top-and-bottom by $(\sec x - 1)$.

Hello AnemoriI think you have a sign wrong. One of the minus signs should be a plus sign. So perhaps you must mean: prove the identity:
$\frac {\sec x - 1}{\tan x } = \frac {\tan x}{\sec x + 1}$
Start with the LHS, and multiply top-and-bottom of the fraction by $(\sec x + 1)$
$\frac {\sec x - 1}{\tan x } =\frac {(\sec x - 1)\color{red}(\sec x + 1)}{\tan x\color{red} (\sec x + 1)}$
$=\frac{\sec^2x-1}{\tan x (\sec x + 1)}$

$=\frac{\tan^2x}{\tan x (\sec x + 1)}$, using the identity $\sec^2x -1 = \tan^2x$

$=\frac{\tan x}{\sec x + 1}$

On the other hand, if it's the other minus sign that's wrong, again you'll start with the LHS, but this time multiply top-and-bottom by $(\sec x - 1)$.