Proofs
- Thread starter lola
- Start date
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
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
Daniel_Feldman
Full Member
- Joined
- Sep 30, 2005
- Messages
- 252
Re: Okay... this is where im getting 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
lola said:
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
- Joined
- Feb 4, 2004
- Messages
- 16,582
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.
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.
Re: Okay... this is where im getting stuck....
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).lola said: