# Identity? issues

• November 11th 2012, 12:23 PM
M670
Identity? issues
Simplify and write the trigonometric expression in terms of sine and cosine:
http://webwork.mathstat.concordia.ca...4c19df41b1.png
http://webwork.mathstat.concordia.ca...14aa0b5111.png?

So I came up with this
http://webwork.mathstat.concordia.ca...485b290d01.png = 1/f(u)

But I am not sure what to do or if I am right?
• November 11th 2012, 12:27 PM
MarkFL
Re: Identity? issues
I believe what you being asked to do is:

$\tan(u)+\cot(u)=\frac{\sin(u)}{\cos(u)}+\frac{\cos (u)}{\sin(u)}=?=\frac{1}{f(u)}$

Where the "?" is, combine terms, and you will get the desired form.
• November 11th 2012, 12:36 PM
M670
Re: Identity? issues
Can you explain what cofunction identities are?

Attachment 25653
• November 11th 2012, 12:44 PM
MarkFL
Re: Identity? issues
Basically, if you have two complementary angles $\alpha$ and $\beta$, i.e.:

$\alpha+\beta=\frac{\pi}{2}$

then a trig. function whose argument is one of the angles will be equal to its co-function evaluated at the other. That is:

$\sin(\alpha)=\cos(\beta)$

$\cos(\alpha)=\sin(\beta)$

$\tan(\alpha)=\cot(\beta)$

$\cot(\alpha)=\tan(\beta)$

$\sec(\alpha)=\csc(\beta)$

$\csc(\alpha)=\sec(\beta)$
• November 11th 2012, 12:45 PM
M670
Re: Identity? issues
Quote:

Originally Posted by MarkFL2
I believe what you being asked to do is:

$\tan(u)+\cot(u)=\frac{\sin(u)}{\cos(u)}+\frac{\cos (u)}{\sin(u)}=?=\frac{1}{f(u)}$

Where the "?" is, combine terms, and you will get the desired form.

Oh does this work out to Sin^2(u)+Cos^2(u) which would equal 1?
• November 11th 2012, 12:47 PM
MarkFL
Re: Identity? issues
Yes, that would be the numerator...what is the common denominator, which is what f(u) is?
• November 11th 2012, 01:01 PM
M670
Re: Identity? issues
Quote:

Originally Posted by MarkFL2
Yes, that would be the numerator...what is the common denominator, which is what f(u) is?

I am not understanding this ? if I found the numerator to be 1 this means 1=1/f(u)?
• November 11th 2012, 01:07 PM
MarkFL
Re: Identity? issues
When you combine the two terms, the numerator simplifies to 1, but there is also the denominator.
• November 11th 2012, 01:20 PM
M670
Re: Identity? issues
I really don't know I am now lost.. I understand what you said, I just can't wrap my mind around this one...
• November 11th 2012, 01:26 PM
skeeter
Re: Identity? issues
$\frac{\sin{u}}{\cos{u}} + \frac{\cos{u}}{\sin{u}} = \frac{\sin^2{u}+\cos^2{u}}{\sin{u}\cos{u}} = \frac{1}{\sin{u}\cos{u}}$
• November 11th 2012, 01:31 PM
M670
Re: Identity? issues
Quote:

Originally Posted by skeeter
$\frac{\sin{u}}{\cos{u}} + \frac{\cos{u}}{\sin{u}} = \frac{\sin^2{u}+\cos^2{u}}{\sin{u}\cos{u}} = \frac{1}{\sin{u}\cos{u}}$

I know my biggest problem is my algebra skill set is very poor right now.... Can you break this down into a few more steps for me to be able to understand even better?
• November 11th 2012, 01:52 PM
skeeter
Re: Identity? issues
$\frac{a}{b} + \frac{b}{a}$

common denominator is $a \cdot b$

$\frac{a \cdot a}{a \cdot b} + \frac{b \cdot b}{a \cdot b}$

$\frac{a^2}{ab} + \frac{b^2}{ab}$

$\frac{a^2+b^2}{ab}$