# Thread: What identity should I use?

1. ## What identity should I use?

What Identity should I use to solve this.

$csc x-\frac{cotx}{secx} =sinx$

2. Originally Posted by mtt0216
What Identity should I use to solve this.

$csc x-\frac{cotx}{secx} =sinx$
You are not solving anything. You are proving an identity. There's a big difference in meaning between these two words.

I suggest you substitute the definitions of csc x, cot x and sec x into the left hand side and simplify the result.

3. I meant to say prove sorry about that.

The problem I have with this problem is that I don't know how to change

$\frac{cotx}{secx}$ into a different form.

Its just the fact thats its in an equation and I just don't know where to start.

4. Originally Posted by mtt0216
I meant to say prove sorry about that.

The problem I have with this problem is that I don't know how to change

$\frac{cotx}{secx}$ into a different form.

Its just the fact thats its in an equation and I just don't know where to start.
Do you know the definition of each in terms of sinx and cosx? It will be in your notes and/or textbook. Then make the substitutions as I said in my first reply.

5. I know the definitions:

csc x = 1/sinx

sec x = 1/cosx

Cot= cosx/sinx

But once I put them into the equation I dont know what to do.

6. Originally Posted by mtt0216
I know the definitions:

csc x = 1/sinx

sec x = 1/cosx

Cot= cosx/sinx

But once I put them into the equation I dont know what to do.
At this level you should be able to do basic algebra on expressions like (a/b)/(c/d) to get (ad)/(bc).

Note: $\frac{a/b}{c/d} = \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = ....$

7. Originally Posted by mr fantastic
At this level you should be able to do basic algebra on expressions like (a/b)/(c/d) to get (ad)/(bc).

Note: $\frac{a/b}{c/d} = \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = ....$
OK sorrryy that I forgot a basic step.......................

8. Originally Posted by mtt0216
I know the definitions:

csc x = 1/sinx

sec x = 1/cosx

Cot= cosx/sinx

But once I put them into the equation I dont know what to do.

$
\frac{1}{\sin{\theta}}
-\frac
{\frac{\cos{\theta}}{\sin{\theta}}}
{\frac{1}{\cos{\theta}}}
\rightarrow
\frac{1}{\sin{\theta}}-\frac{\cos{\theta}}{\sin{\theta}}
\frac{\cos{\theta}}{1}
\rightarrow
\frac{1-\cos^2{\theta}}{\sin{\theta}}
\rightarrow
\frac{\sin^2{\theta}}{\sin{\theta}}= \sin{\theta}
$

9. Yea thats what I had. thanks