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



Please verify if you can!!

There are always many methods to tackle these critters, but I will do a somewhat lengthy one so you can see the identities. OK?

\frac{cot(x)}{1-tan(x)}+\frac{tan(x)}{1-cot(x)}

Like any addition, you must make the denominators the same:

Let's multiply the right side by -tan(x):

\frac{cot(x)}{1-tan(x)}+\frac{tan(x)}{1-cot(x)}\cdot\frac{-tan(x)}{-tan(x)}

Now you have:

\frac{cot(x)}{1-tan(x)}-\frac{tan(x)}{1-tan(x)}

=\frac{cot(x)-tan^{2}(x)}{1-tan(x)}

Using cot(x)=\frac{cos(x)}{sin(x)} \;\ and \;\ tan(x)=\frac{sin(x)}{cos(x)}:

We simplify it down to:

\frac{cos^{2}(x)+sin(x)cos(x)+sin^{2}(x)}{sin(x)co s(x)}

Since cos^{2}(x)+sin^{2}(x)=1

\frac{1+sin(x)cos(x)}{sin(x)cos(x)}

So, we're getting there:

\frac{1}{sin(x)cos(x)}+\frac{sin(x)cos(x)}{sin(x)c os(x)}

=\frac{1}{sin(x)cos(x)}+1

Now, \frac{1}{sin(x)cos(x)}=cot(x)+tan(x)

Therefore, we have... drum roll please:

cot(x)+tan(x)+1


This is a rather lengthy derivation but if you can follow it, you can follow amore simplistic method.

I appreciate the reply but I still am confused by some of your answer. First I don't see how you derived the new equation when you put it all in terms of sin and cos. Then I don't see how 1/sin(x)cos(x)can be equal to cot(x)+tan(x). If you could elaborate more I would appreciate it.

That's how you do these things a lot of the time is by converting to sine and cosine.

You must know variuos identities in order to transform the left side into the right side.

cot(x)=\frac{cos(x)}{sin(x)}
tan(x)=\frac{sin(x)}{cos(x)}

So, we have:

\frac{cos(x)}{sin(x)}+\frac{sin(x)}{cos(x)}

Cross multiply:

\frac{cos^{2}(x)+sin^{2}(x)}{sin(x)cos(x)}

Remember, cos^{2}(x)+sin^{2}(x)=1

Therefore:

\frac{1}{sin(x)cos(x)}=cot(x)+tan(x)

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



Please verify if you can!!!!
I have no idea

I wish I knew who you were. I have mrs.philips as my precal teacher