Question

(cscX-1)/(cotX)=(cotX)/(cscX+1)

Answer 1

put \(a=\cos(x), b=\sin(x)\) and this becomes
\[\frac{\frac{1}{b}-1}{\frac{a}{b}}=\frac{\frac{a}{b}}{\frac{1}{b}+1}\]

Answer 2

ok, I'm following. Let me work on it a sec...

Answer 3

Ok, nope. I'm completely lost. I never should have taken this class online!
so far I am ending up with (1/sinX)-1=((1/sinX)+1)*(cos^2X/sin^2X) how am I doing so far?

Answer 4

\[\frac{\frac{1}{b}-1}{\frac{a}{b}}=\frac{\frac{a}{b}}{\frac{1}{b}+1}\] multiply by \(b\) top and bottom to get rid of the compound fraction to get
\[\frac{1-b}{a}=\frac{a}{1-b}\]

Answer 5

then cross multiply to get \[(1+b)(1-b)=a^2\] or
\[1-b^2=a^2\]

Answer 6

since \(a=\cos(x),b=\sin(x)\) this is saying that
\[1-\sin^2(x)=\cos^2(x)\] which you know is true, since
\[\cos^2(x)+\sin^2(x)=1\]
work the algebra backwards and you have your "proof"

Answer 7

Awesome! Thank you! I am going to work through it all on my own to see if I understand it. So far I've just caught up to getting where the very first part came from. I am old and a bit rusty on my fundamentals.

Answer 8

I am getting stuck on the part where you are multiplying b with (1/b)-1 or (1/b)+1 why wouldn't the b's cancel each other out?

Answer 9

We know that \(\cot^2(X) + 1 = \csc^2(X) \)\[\Rightarrow \cot^2(X) = \csc^2(X) - 1 \]\[\Rightarrow \cot^2(X) = (\csc(X) + 1)(\csc (X) - 1) \]\[\Rightarrow \dfrac{\cot^2(X)}{\csc(X) + 1} = \csc(X) - 1\]\[\Rightarrow \dfrac{\cot(X)}{\csc(X) + 1} = \dfrac{\csc(X)-1}{\cot(X)}\]