Proofs

[lola]

HI... I hope there is someone that can help me..... I am really confused. I need to prove that

(sec x-tan x)(csc x+1)=cot x

I cant seem to prove it but I remember that the teacher said it was true. I honestly don't think it is..... What about u :?:

 

[stapel]

the teacher said it was true. I honestly don't think it is.

On what basis do you think this?

Eliz.

 

[lola]

Okay... this is where im getting stuck....

Do u see what im doing wrong :?:
or how im getting stuck....



so we have:

(sec x - tan x)(csc x +1) = cot x

then i get :
(1/cos x - sin x/cos x)(1/sin x + 1) = cos x/sin x

so then I combine like terms:
(1-sin x/cos x)(1+sin x/sin x)

and now im stuck

 

[lola]

this is what I think

I can't seem to prove it. Thats why... but in class i can specifically remember the teacher giving a clue to the first one - She said it was possible but I don't see it.

Do you think you can help

 

[Daniel_Feldman]

Re: Okay... this is where im getting stuck....

Do u see what im doing wrong :?:
or how im getting stuck....



so we have:

(sec x - tan x)(csc x +1) = cot x

then i get :
(1/cos x - sin x/cos x)(1/sin x + 1) = cos x/sin x

so then I combine like terms:
(1-sin x/cos x)(1+sin x/sin x)

and now im stuck

You have

(1-sin x/cos x)(1+sin x/sin x)

=(1-sinx)(1+sinx)/sinxcosx

=(1-sin^2(x))/sinxcosx

Now, remember the identity


sin^2(x)+cos^2(x)=1

So 1-sin^2(x)=cos^2(x)

So back to the problem

(1-sin^2(x))/sinxcosx

=cos^2(x)/sinxcosx

=(cosxcosx)/(sinxcosx)

Cosines cancel, and we're left with

cosx/sinx

=cotx

 

[stapel]

Note: Grouping symbols will serve to clarify your meaning greatly. I think I know what you mean, but if I'm right, you don't mean what you typed. For instance, "1 + sinx/sinx" means "1 + 1" or "2", but I'm betting you mean "[1 + sin(x)] / sin(x)".

You have gotten this far:


. . . . .$\displaystyle \large{\left(\frac{1}{\cos{(x)}}\,- \,\frac{\sin{(x)}}{\cos{(x)}}\right)\, \left(\frac{1}{\sin{(x)}}\,+\,1\right)\,=\,\frac{\cos{(x)}}{\sin{(x)}}}$


. . . . .$\displaystyle \large{\left(\frac{1\,-\,\sin{(x)}}{\cos{(x)}}\right)\; \left(\frac{1\,+\,\sin{(x)}}{\sin{(x)}}\right)\,= \,\frac{\cos{(x)}}{\sin{(x)}}}$


Now multiply through the numerators on the left-hand side:


. . . . .$\displaystyle \large{\frac{1\,-\,\sin^2{(x)}}{\cos{(x)}\sin{(x)}}\,= \,\frac{\cos{(x)}}{\sin{(x)}}}$


Simplify that numerator, cancel, and see what you end up with.

Eliz.

 

[lola]

THANKS

THANK YO SO MUCH :!: :!: :!: :!:

 

[lola]

I see

I see what you mean... i guess i did forget about the parenthesis.... Thanks again for your help stapel

 

[Unco]

Re: Okay... this is where im getting stuck....

so we have:

(sec x - tan x)(csc x +1) = cot x

then i get :
(1/cos x - sin x/cos x)(1/sin x + 1) = cos x/sin x

It is a bit more rigorous (and possibly more preferable for your examiner) to deal with the LHS and RHS separately, as, technically, we do not if the original equation is true (until we prove it).