Math Help - Trig identities! Help please.

I need to reduce trig identities to the simplest form. i was wondering if you could take me further.

(secx)(cscx)-tanx
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cotx(1/secx)+sinx

i changed the top to (1/cosx)(1/sinx)-(sinx/cosx)

Thats as far as i can do it

2. Keep going ... Convert everything into terms of sinx and cosx and combine the fractions by finding the common denominator. For example in the numerator:

$\sec x \csc x - \tan x$

$= \frac{1}{\cos x \sin x} - \frac{\sin x}{\cos x}$

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

$= \frac{\overbrace{1 - \sin^{2} x}^{\mbox{Look familiar?}}}{\cos x \sin x}$

3. so i get :

$(cos^2x)/(cosxsinx)$

for the numerator?

4. Hello, Airjunkie!

Reduce to the simplest form: . $\frac{\sec x\csc x -\tan x}{\cot x\!\cdot\!\frac{1}{\sec x}+\sin x}$

i changed the top to: . $\frac{1}{\cos x}\!\cdot\!\frac{1}{\sin x} -\frac{\sin x}{\cos x}$ . . . . Good!
Get a common denominator and combine the fractions.

. . $\frac{1}{\cos x\sin x} - \frac{\sin x}{\cos x}\cdot\frac{\sin x}{\sin x} \;\;=\;\;\frac{1}{\cos x\sin x} - \frac{\sin^2\!x}{\cos x\sin x}$

. . $= \;\;\frac{1-\sin^2\!x}{\cos x\sin x} \;\;=\;\;\frac{\cos^2\!x}{\cos x\sin x} \;\;=\;\;\frac{\cos x}{\sin x}$

The denominator is: . $\frac{\cos x}{\sin x}\!\cdot\!\frac{\cos x}{1} + \sin x \;\;=\;\;\frac{\cos^2\!x}{\sin x} + \frac{\sin^2\!x}{\sin x}$
. . $=\;\;\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}}}{\sin x} \;\;=\;\;\frac{1}{\sin x}$

The problem becomes: . $\frac{\dfrac{\cos x}{\sin x}}{\dfrac{1}{\sin x}} \;\;=\;\;\boxed{\cos x}$

5. Wow, i see what i did wrong. I understand it better. But not I have to solve a much more complicated one, and im quite confused still.

i didn't fall for the 1/1 at least :P

6. Let's concentrate piece by piece.

1. $\frac{\cos \left(\frac{\pi}{2} - \theta\right)}{sin^{2}\left(\frac{\pi}{2} - \theta\right) + cos^{2}\left(\frac{\pi}{2} - \theta\right)}$

First off, you should immediately recognize the identity found in the denominator. As for the numerator, if you know your sin
and cos graphs like the back of your hand:
$\cos \left(\frac{\pi}{2} - \theta \right) = \overbrace{\cos \left(\theta - \frac{\pi}{2}\right)}^{ \cos (- x ) = \cos x }$

which is just y = cos x shifted to the right by pi/2. What would that be?

2. $\frac{\frac{1}{\frac{1}{\sin\theta}}}{\sec \theta} = \frac{\sin \theta}{\frac{1}{\cos \theta}} = \mbox{etc.}$

3. $- \frac{1}{\cot (\pi + \theta)} = - \tan(\theta + \pi)$
Ignoring the negative sign for now, if you shift the graph y = tanx to the left by pi, what do you get?

4.
$\sin^{2}(- \theta) + \cos^{2}(\theta) = (\sin \left(-\theta\right))^{2} + \cos^{2}(\theta) = \underbrace{(-\sin (\theta))^{2}}_{\sin(-\theta) = -\sin (\theta)} + \cos^{2} (\theta) = \sin^{2}\theta + \cos^{2} \theta$

7. Im not trying to be mean but unless i see steps it cant help me.

By the way this isnt really for a grade, but she said these will be on the tests. So id really like to study and memorize the rules with them. I need to see the whole thing. Ill try to look over it again