Hey guys,
How would I go about solving
$\displaystyle \frac{cos(2x)}{1-sin(2x)}=\frac{1+tan(x)}{1-tan(x)}$
I have already tried cross-multiplying, but that didn't work out too well
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
$\displaystyle \frac{1+tan(x)}{1-tan(x)}$
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 ...
$\displaystyle \frac{\cos(2x)}{1 - \sin(2x)} =$
$\displaystyle \frac{\cos^2{x} - \sin^2{x}}{1 - 2\sin{x}\cos{x}} =$
$\displaystyle \frac{\cos^2{x} - \sin^2{x}}{\cos^2{x} - 2\sin{x}\cos{x} + \sin^2{x}} =$
$\displaystyle \frac{(\cos{x}-\sin{x})(\cos{x}+\sin{x})}{(\cos{x} - \sin{x})^2} =$
$\displaystyle \frac{\cos{x}+\sin{x}}{\cos{x}-\sin{x}} =$
divide every term by $\displaystyle \cos{x}$ ...
$\displaystyle \frac{1+\tan{x}}{1-\tan{x}}$
someone else will probably show you how to work it from the right side and transform it to the left.
I have another question, I think im further in this one:
$\displaystyle sec(x) + tan(x) = \frac{sin(x)}{1-sin(x)}$
I'm trying to turn the left side into the right:
$\displaystyle =\frac{1}{cos(x)}+tan(x)$
$\displaystyle =\frac{1+tan(x)cos(x)}{cos(x)}$
$\displaystyle =\frac{1+sin(x)}{cos(x)}$
Once i get there, I get stuck and don't know what to do next. Any helP?
As skeeter showed, it's not an identity.
However, if the question is.... "Solve for x", then
$\displaystyle \displaystyle\frac{1+Sin(x)}{Cos(x)}=\frac{Sin(x)} {1-Sin(x)}\Rightarrow\ 1-Sin^2(x)=Sin(x)Cos(x)$
$\displaystyle \Rightarrow\ Sin(x)Cos(x)=Cos^2(x)\Rightarrow\ Cos(x)[Cos(x)-Sin(x)]=0$
$\displaystyle Cos(x)=0,\ Cos(x)=Sin(x)$
Solve for x.
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.