1. ## Trig Identity Simplification

Hey guys,

How would I go about solving
$\frac{cos(2x)}{1-sin(2x)}=\frac{1+tan(x)}{1-tan(x)}$

I have already tried cross-multiplying, but that didn't work out too well

2. sorry guys its not solving, and my first steps were wrong. Im just trying to prove it true. so, I have to manipulate one side to look like the other. I have expanded cos(2x) and sin(2x) with the double angle identities, but still can't get it to look like
$\frac{1+tan(x)}{1-tan(x)}$

3. Originally Posted by spycrab
Hey guys,

How would I go about solving
$\frac{cos(2x)}{1-sin(2x)}=\frac{1+tan(x)}{1-tan(x)}$

I have already tried cross-multiplying, but that didn't work out too well
one does not "cross-multiply" to prove an identity. work with one side and transform it to the other side.

working on the left side ...

$\frac{\cos(2x)}{1 - \sin(2x)} =$

$\frac{\cos^2{x} - \sin^2{x}}{1 - 2\sin{x}\cos{x}} =$

$\frac{\cos^2{x} - \sin^2{x}}{\cos^2{x} - 2\sin{x}\cos{x} + \sin^2{x}} =$

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

$\frac{\cos{x}+\sin{x}}{\cos{x}-\sin{x}} =$

divide every term by $\cos{x}$ ...

$\frac{1+\tan{x}}{1-\tan{x}}$

someone else will probably show you how to work it from the right side and transform it to the left.

4. ## another question

I have another question, I think im further in this one:
$sec(x) + tan(x) = \frac{sin(x)}{1-sin(x)}$
I'm trying to turn the left side into the right:
$=\frac{1}{cos(x)}+tan(x)$
$=\frac{1+tan(x)cos(x)}{cos(x)}$
$=\frac{1+sin(x)}{cos(x)}$

Once i get there, I get stuck and don't know what to do next. Any helP?

5. $\displaystyle \sec{x} + \tan{x} = \frac{\sin{x}}{1-\sin{x}}$

not an identity ... let x = 0 and you'll see the left side does not equal the right side.

6. Ahh. well we weren't allowed to use our calculator on this assignment so I never bothered to try it. Thanks for helping me not waste my time

7. you should be able to evaluate trig function values w/o a calculator using unit circle values.

8. Originally Posted by spycrab
I have another question, I think im further in this one:

$sec(x) + tan(x) = \frac{sin(x)}{1-sin(x)}$

I'm trying to turn the left side into the right:

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

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

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

Once i get there, I get stuck and don't know what to do next. Any helP?
Should that have been

$Sec(x)+Tan(x)=\displaystyle\frac{Cos(x)}{1-Sin(x)}$ ?

since...

$\displaystyle\frac{1+Sin(x)}{Cos(x)}=\frac{Cos(x)} {1-Sin(x)}$

as $[1+Sin(x)][1-Sin(x)]=1-Sin^2(x)=Cos^2(x)$

9. Thanks, but no it wasn't. I just read it again. Thanks!

10. Originally Posted by spycrab
Thanks, but no it wasn't. I just read it again. Thanks!
As skeeter showed, it's not an identity.

However, if the question is.... "Solve for x", then

$\displaystyle\frac{1+Sin(x)}{Cos(x)}=\frac{Sin(x)} {1-Sin(x)}\Rightarrow\ 1-Sin^2(x)=Sin(x)Cos(x)$

$\Rightarrow\ Sin(x)Cos(x)=Cos^2(x)\Rightarrow\ Cos(x)[Cos(x)-Sin(x)]=0$

$Cos(x)=0,\ Cos(x)=Sin(x)$

Solve for x.

11. Originally Posted by spycrab
Hey guys,

How would I go about solving

$\frac{cos(2x)}{1-sin(2x)}=\frac{1+tan(x)}{1-tan(x)}$

I have already tried cross-multiplying, but that didn't work out too well
For this question, you can show that both sides are equal for all x.
Generally you would be asked to prove that both sides are the same.

Solving, on the other hand generally means that the two sides are equal for some specific value(s) of x.
The two sides are not equal in general, but we can look for the value(s) of x for which they are equal.