Why Can't I Simplify This Trigonometric Equation?

[Adam2987]

Homework Statement

1/cscx-sinx = secx tanx

Homework Equations

cscx = 1/sinx
secx = 1/cosx

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?

 

[rock.freak667]

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?

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

The Attempt at a Solution

1/cscx-sinx = secx tanx

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

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

 

[Adam2987]

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

 

[rock.freak667]

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

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

Redo it.

 

[Adam2987]

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.

 

[rock.freak667]

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.

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

 

[Adam2987]

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

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

 

[Adam2987]

err or would it be sinx/sinx-sinx?

 

[physicsman2]

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

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