I got stumped on this trig proof

[cosx + sinx]/cosx + [cosx-sinx]/cosx = cscxsecx








I converted Csc(x)Sec(x) to 1/sin(x)cos(x) but I just don't know where to begin with the left side.

This post was edited by Jwoww on Dec 12 2013 12:15am

Find a common denominator on the left side, then you can combine all the terms.
If it's set up like I think it is, things should cancel.
I also recommend using parenthesis or brackets to distinguish between your fractions

Alright I'll give it a shot!

Alright so I multiplied the left side by sinx/sinx to get a common denominator of sinxcosx

The sin^2x cancel each other since (+/-) but

what does sinxcosx + sinxcosx =?

Is that the same as sin^2xcos^2x? If so then it would equal 1 and give me the left being 1/sinxcosx and the right as 1/sinxcosx

but I'm not entirely sure if that step is right.

sinxcosx + sinxcosx = 2sinxcosx

using the trig identity, you can turn that into sin(2x)

also, do you mind writing some parenthesis in? i cant tell what the equation is

Is it [cosx + sinx]/cosx + [cosx-sinx]/cosx
or
cosx + (sinx/cosx) + cosx - (sinx/cosx)

just bold the right one

My bad! It's the first one you posted

Been so frustrated with math lately, not very good at it but trying to understand it all.

Hmm, since they have the same denominator, you can just combine the terms on the left side and simplify. You get 2cos(x) / cos(x)
But after doing that, it simplifies to:
2 = csc(x)sec(x)

I may not have the original equation correct.

E: i'm long overdue for some sleep, time for me to hit the sack! But there's plenty of excellent math people here to help. Good luck!

This post was edited by TritonV8 on Dec 12 2013 12:24am

same thing i'm getting...u sure that's the right equation, Jwoww?