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this is the first time I've asked a question here, so bare with me...

I'm in Year 12 Maths B (kinda like Maths Extension) and, though we have not been told anything at all whatsoever about Cosecant, Secant and Cotangent, I got curious. Please excuse my comparatively limited knowledge of mathematics.

So, I had to differentiate $\frac{1}{\tan(x)}$ into it's simplest form, and I have no source of answers to check with, so I used Wolfram|Alpha. It told me that the simplest form was $\csc^2(x)$ (by the way, sorry if I haven't yet discovered how to make maths look proper on this page) which got me confused. After many research, I discovered that these other three trigonometric functions are the reciprocals of the major three. So, working through my problem, I got to this point:

$$-\sec^2(x) \cot^2(x) = -\csc^2(x)$$

This is where my understanding fails. What processes exist between the two equations immediately above this text? Why does $$-\sec^2(x) \cot^2(x) = -\csc^2(x)$$

A detailed, step-by-step instruction on how/why this is what it is would really help me understand, and would be much appreciated.

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    $\begingroup$ $$\sec(x) = \frac{1}{\cos(x)}$$ $$\cot(x) = \frac{\cos(x)}{\sin(x)}$$ $$\csc(x) = \frac{1}{\sin(x)}$$ $\endgroup$
    – Luigi D.
    Mar 4, 2015 at 9:00
  • $\begingroup$ Yes, I get this, but how does -sec(x)^2 * cot(x)^2 = -(1/sin(x))^2 $\endgroup$
    – Justin
    Mar 4, 2015 at 9:03
  • $\begingroup$ Given what Luigi has told you. Why dont you put it in your equation and see for yourself. Check for LHS=RHS. $\endgroup$
    – MonK
    Mar 4, 2015 at 9:05
  • $\begingroup$ $\frac{1}{\cos(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{1}{\sin(x)}$. Square both sides and add the minus sign. $\endgroup$
    – Luigi D.
    Mar 4, 2015 at 9:05
  • $\begingroup$ :3 sorry if I caused any frustration, however mild. I understand now, thanks for explaining that. +1, or whatever you guys get on stack exchange. $\endgroup$
    – Justin
    Mar 4, 2015 at 9:07

1 Answer 1

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Welcome to MSE.

$$-\sec^2(x)\cot^2(x)=-\frac{1}{\cos^2(x)}\frac{\cos^2(x)}{\sin^2(x)}=-\frac{1}{\sin^2(x)}=-\csc^2(x)$$

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    $\begingroup$ To produce $\sin x$, type \sin x between dollar signs. $\endgroup$
    – N. F. Taussig
    Mar 4, 2015 at 9:54
  • $\begingroup$ ^I don't see the necessity though..... Why bother? It just looks a bit italicized I guess....But I'll do so from now. $\endgroup$
    – Panglossian Oporopolist
    Mar 4, 2015 at 10:02

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