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I need to show two auxiliary trigonometric identities:

1) $\sec^2x = \tan ^2x + 1 (\cos x \neq 0)$

2) $\csc^2x = \cot^2x +1 (\sin x \neq 0)$

How could I do it?

[Original Portuguese]

Identidades Trigonométricas Auxiliares Sendo preciso mostrar a demonstração das duas fórmulas das IDENTIDADES TRIGONOMÉTRICAS AUXILIARES:

1) sec²x = tg²x + 1 (cos x ≠ 0)

2) cossec²x = cotg²x = 1 (sen x ≠ 0)

Como seria possível fazê-lo???

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Welcome! It will be nice if you write in English . – Iuli Sep 11 '12 at 21:47
@Iuli Consensus on meta in the past has been very strongly in favor of people posting questions in their native languages. The idea that it is much better to get a coherent, well-expressed question in Portuguese, and which can be translated, than to get an unintelligible question in English. – MJD Sep 11 '12 at 22:14
@ MJD: I translated the question to English already, it's just that the OP has not yet accepted the edit. – Rod Carvalho Sep 11 '12 at 22:18
@RodCarvalho Are you sure? I would expect that your edit would go into the queue to be approved by high-reputation users, not by OP. But it is not in the queue. – MJD Sep 11 '12 at 22:28
@ MJD: It said it was sent to the queue after I submitted the edit. In any case, Peter translated it in the meantime. – Rod Carvalho Sep 11 '12 at 22:32

Starting from $\sin^2 x + \cos^2 x = 1$, you get, dividing by $\cos^2 x$ (if this is $\ne 0$) \[ \frac{\sin^2x}{\cos^2 x} + \frac{\cos^2 x}{\cos^2 x} = \frac 1{\cos^2 x} \] which gives \[ \tan^2 x + 1 = \mathrm{sec}^2 x \] Dividing the first identity by $\sin^2 x$ we get \[ \frac{\sin^2x}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x} = \frac 1{\sin^2 x} \] which gives \[ 1 + \cot^2 x = \mathrm{csc}^2 x \]

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Oh my God, thanks!!! Really really really really thank you!! I was not finding it in any brazilian website, you've helped a lot. FOREVER GRATIFIED!!! I'm so happy for this. Sorry for bother you...Haha – Ana Sep 11 '12 at 23:35

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