# Solving Trig Identities.

1. Apr 26, 2010

1. The problem statement, all variables and given/known data

1/cscx-sinx = secx tanx

2. Relevant equations

cscx = 1/sinx
secx = 1/cosx

3. The attempt at a solution

1/cscx-sinx = secx tanx

L.S.
= 1/cscx-sinx
= 1/(1/sinx)-sinx

R.S.
= secx tanx
= (1/cosx)(sinx/cosx)

This is where I'm getting confused. Why can't I make the L.S equal the Right side?

2. Apr 26, 2010

### rock.freak667

Start with one side alone and make that match the other side.

What happens if you multiply both the numerator and denominator by sinx/sinx ?

3. Apr 26, 2010

(1/sinx)(sinx/sinx) = sin^2x - sinx?

4. Apr 26, 2010

### rock.freak667

$$\frac{1}{\frac{1}{sinx}-sinx} \times \frac{sinx}{sinx}$$

Redo it.

5. Apr 26, 2010

Hmmm. Do I multiply everything in the first denominator by sinx? Or just the - sinx?

I get sinx/sinx-sin2x. I think... I've never seen mulitiplication like this. It's probably something easy, I've just never done it yet.

6. Apr 26, 2010

### rock.freak667

Multiply everything in the numerator by sinx, and multiply everything in the denominator by sinx.

7. Apr 26, 2010

ok so I get sinx/(sinx/sin^2x)-sin^2x

If I divide that I end up with sinx-sinx = 0?

8. Apr 26, 2010

err or would it be sinx/sinx-sinx?

9. Apr 26, 2010

### physicsman2

If that's the actual problem(can't tell from your original post), then:

get a common denominator:

1/((1-((sin x)^2))/ sin x) becomes:

(sin x)/(1-((sin x)^2)) --> 1 - sin^2 x = cos^2 x:

(sin x)/((cos x)^2) = sec x tan x