Why Can't I Simplify This Trigonometric Equation?

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Adam2987
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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?
 
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Adam2987 said:

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.


Adam2987 said:

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 ?
 
(1/sinx)(sinx/sinx) = sin^2x - sinx?
 
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.
 
Adam2987 said:
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.
 
ok so I get sinx/(sinx/sin^2x)-sin^2x

If I divide that I end up with sinx-sinx = 0?
 
err or would it be sinx/sinx-sinx?
 
rock.freak667 said:
[tex]\frac{1}{\frac{1}{sinx}-sinx} \times \frac{sinx}{sinx}[/tex]

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