Trouble in pre-cal, proofs

1. Mar 9, 2004

BWL

I know this is below most of those that peruse these forums, but I've been giving myself an ulcer trying to figure these silly things out.

The first problem starts out as
(1-sin^2(x))(1+tan^2(x))=1
and I've got it down to
(sin^2(x)/tan^2(x))-sin^2(x)=1
but from there I've got no idea.

The second was
(1+cot(x))/csc(x)=(1+tan(x))/sec(x)
and I've got the right side down to
(cos/1)+(1/csc)
but how that becomes (1+cot(x))/csc(x)is beyond me.

2. Mar 9, 2004

Wooh

For the first: it becomes cos^2(x)sec^2(x)=1, and since cos(x)=1/sec(x) you get 1=1 and it checks out. (1+tan^2(x)=sec^2(x), and 1-sin^2(x)=cos^2(x)

Ok, for the second, first we simplify both sides (on their own! we don't relate them, but merely simplify)
(1+cos/sin)sin=(1+sin/cos)cos (I have dropped the (x) for easy of typing, you get what I mean though)
distributing on both sides yields
sin+cos=cos+sin

It is important to note that you can manipulate one side to a point, and then manipulate the other to that same point and that is completely valid! Just make sure to manipulate within itself! To do otherwise is to assume they are equal, thus destroying the purpose of the proof in the first place.

3. Mar 9, 2004

Janitor

Possible solution to the first problem.

Are you trying to prove that your first equation is an identity?

If so, note that since sin^2(x)+cos^2(x)=1, you have that 1-sin^2(x) = cos^2(x). Then note that since tan(x)=[sin(x)]/cos(x), you have that tan^2(x)=[sin^2(x)]/cos^2(x), so that 1+tan^2(x)=1+[sin^2(x)]/cos^2(x)=[cos^2(x) + sin^2(x)]/cos^2(x)=1/cos^2(x).

So the product of the two terms is cos^2(x) times 1/cos^2(x) which is 1, as was to be proved.

4. Mar 9, 2004

BWL

Thanks for the help, really. The teacher decided giving us a unit test on this stuff the day we got back from spring break was a good idea.