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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.

N. F. Taussig
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Justin
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    $$\sec(x) = \frac{1}{\cos(x)}$$ $$\cot(x) = \frac{\cos(x)}{\sin(x)}$$ $$\csc(x) = \frac{1}{\sin(x)}$$ – Luigi D. Mar 04 '15 at 09:00
  • Yes, I get this, but how does -sec(x)^2 * cot(x)^2 = -(1/sin(x))^2 – Justin Mar 04 '15 at 09:03
  • Given what Luigi has told you. Why dont you put it in your equation and see for yourself. Check for LHS=RHS. – MonK Mar 04 '15 at 09:05
  • $\frac{1}{\cos(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{1}{\sin(x)}$. Square both sides and add the minus sign. – Luigi D. Mar 04 '15 at 09:05
  • :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. – Justin Mar 04 '15 at 09:07

1 Answers1

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Welcome to MSE.

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

N. F. Taussig
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Panglossian Oporopolist
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