Help plz

[Anaf]

I got this trigonometric problem help me

 

[BigBeachBanana]

I got this trigonometric problem help me

Divide the first equation by [imath]\frac{1}{\cos^2(A)}[/imath].

 

[Otis]

It also helps to remember:

[imath]\text{tan}(A) = \frac{1}{\text{cot}(A)}[/imath]

[imath]\text{cot}(A) = \frac{1}{\text{tan}(A)}[/imath]



[imath]\;[/imath]

 

[The Highlander]

It also helps to remember:

[imath]\text{cot}(A) = \frac{1}{\text{cos}(A)}[/imath]

[imath]\text{cos}(A) = \frac{1}{\text{cot}(A)}[/imath]



[imath]\;[/imath]

Hmmm, ?

That's funny (funny peculiar, not funny Ha Ha). In the school I went to we were taught that:

[imath]\text{cot}(A) = \frac{1}{\text{tan}(A)}[/imath]
and
[imath]\text{tan}(A) = \frac{1}{\text{cot}(A)}[/imath]

Is this an American convention???

 

[Otis]

Sigh. That is what sometimes happens when I'm rushed to fix stubborn typos, and I get kicked off the site before I can finish. (I've often wondered whether submitting a quick succession of edits to the same post confuses the server.) Most irritating.

?

[imath]\;[/imath]

 

[lex]

[imath]\cot(A)=\tfrac{\cos(A)}{\sin(A)} \hspace5ex \text{(when the expression is defined)}\\ \text{ }\\ \cot(A)≠\tfrac{1}{\tan(A)} \text{ when }A=\tfrac{(2n+1)\pi}{2}, \hspace3ex \text{the expression on the right-hand side being undefined. Rather } \cot(A)=\tfrac{\cos(A)}{\sin(A)}=0\text{ then.} [/imath]

 

[The Highlander]

[imath]\cot(A)=\tfrac{\cos(A)}{\sin(A)} \hspace5ex \text{(when the expression is defined)}\\ \text{ }\\ \cot(A)≠\tfrac{1}{\tan(A)} \text{ when }A=\tfrac{(2n+1)\pi}{2}, \hspace3ex \text{the expression on the right-hand side being undefined. Rather } \cot(A)=\tfrac{\cos(A)}{\sin(A)}=0\text{ then.} [/imath]

Thanks, lex, I don't know what we'd do without your help to keep us right. ?

 

[Otis]

Yes. Thanks for reminding folks who don't always remember the underlined part:

Identities are statements (equations) that are true for any value of the unknown, when both sides are defined.



[imath]\;[/imath]

 

[lex]

Thanks, lex, I don't know what we'd do without your help to keep us right. ?

Just thought I'd throw another spanner in the works!!

 

[lex]

Yes. Thanks, lex, for reminding folks who don't always remember the underlined part:

Identities are statements (equations) that are true for any value of the unknown, when both sides are defined.

Interesting. I hadn't been told. I always believed the equality broke down too when one side was defined and the other not.

 

[Otis]

I always believed the equality broke down too when one side [is] defined and the other [is] not

It does. If only one side is defined, then both sides are not defined.

[imath]\;[/imath]

 

[lex]

If only one side is defined, then both sides are not defined.

I assume you mean that in that case, it is not true that both sides are defined.
I suppose I am uncomfortable with the identity: [imath]\sqrt{-x^2}\equiv x[/imath]

...de gustibus...and all that.