I need help Proving an Identity

Whoppie

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Nov 14, 2006
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I need some guidance on proving these identitles. Any help would be appreciated.

1) csc x cot x/sec x = cot^2x

2) csc A(sin^2 A +cos^2 A tan A)/sin A+cos A =1
 
#1:

\(\displaystyle \L\\\frac{csc(x)cot(x)}{sec(x)}\\=\frac{\frac{1}{sin(x)}\cdot\frac{1}{tan(x)}}{\frac{1}{cos(x)}}\\=\left(\frac{1}{sin(x)}\cdot\frac{1}{tan(x)}\right)cos(x)\\=\frac{cos(x)}{sin(x)}\cdot\frac{1}{tan(x)}\)

Now, see it?.
 
Oh my god I didn't even notice that. Because Cos(x)/Sin(x) is Cot(x). I was doing something totally different and I kept getting stuck Thanks alot.
 
For the second one would I have to factor out (sin^2 A +cos^2 A tan A) with foil and then start work on it or is there another way?
 
What do you mean "factor with foil"? Why not just "factor"? Anyway, don't do that.

csc(x) = 1/sin(x)

tan(x) = sin(x)/cos(x)

That's about all you need. The rest is algebra.
 
Sorry I think I said that backwards I shudn't have put foil in there at all
 
Absolutely not! Try it:

(2^2 + 3^2)/(2 + 3) = (4+9)/5 = 13/5

After your proposed reduction: 2+3 = 5 -- Nope! That's not 13/5.

Why do you still have sine squared and cosine squared? You didn't use your hints.

Simplify and factor the numerator AFTER using both hints.
 
Sorry I'm all confused would the Sin^2A+Cos^2A become 1 or Sin^2A+Cos^2x+CosxSinx. Sorry again I always make things harder then they shud be and confuse myself
 
Alas, poor Whoppie, you are making it harder than it needs to be.

Remember, cscA=1sinA\displaystyle cscA=\frac{1}{sinA}.

\(\displaystyle \L\\\frac{\frac{1}{sinA}(sin^{2}A+cos^{2}AtanA)}{sinA+cosA}\)



\(\displaystyle \L\\\frac{\frac{sin^{2}A}{sinA}+\frac{cos^{2}AtanA}{sinA}}{sinA+cosA}\)

Now, simplify the top and you should see it.

Remember what \(\displaystyle \L\\\frac{tanA}{sinA}\) is equivalent to?.
 
Yea I was doing something different and it was making me mad. For some reason I was trying to take the Tan out completely. I don't know when it comes to math I'm stumped for the most part just gotta keep trying I guess.

Thank u both for all your help
 
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