Rationalizing $\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}$ in two ways gives different answers I have a doubt see when we rationalize denominator of expression $$\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}- \sqrt{1-\cos x}}$$
we get answer $$\frac{1+\sin x}{\cos x}$$ but when we rationalize numerator we get
$$\frac{\cos x}{1+\sin x}$$
How is this possible, because rationalizing means just multiplying by $1$?
 A: You might have an error, check that at the end you should have that rationalazing the denominator you should have that $$\cfrac{1+\sin{x}}{\cos{x}},$$ and the numerator you should have that $$\cfrac{\cos{x}}{1-\sin{x}},$$ wich it´s always the same by $(\cos{x})^2+(\sin{x})^2=1$
A: Putting $a$ for $\cos x$,
you have
(notice the MathJax)
$\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}
$
I now do the rationalizing
(using
$(u+v)(u-v)=u^2-v^2)$
and later substitute
$a = \cos(x)$
and use
$\sin^2(x)+\cos^2(x) = 1$.
I am doing this
in excruciating detail
so you can see
all the steps involved.
Once you understand these,
you should be able
to do this kind of thing
by yourself.
$\begin{array}\\
\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}
&=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}+\sqrt{1-a}}\\
&=\dfrac{1+a+2\sqrt{1+a}\sqrt{1-a}+1-a}{(1+a)-(1-a)}\\
&=\dfrac{2+2\sqrt{(1+a)(1-a)}}{2a}\\
&=\dfrac{1+\sqrt{1-a^2}}{a}\\
&=\dfrac{1+\sqrt{1-\cos^2(x)}}{\cos(x)}\\
&=\dfrac{1+\sin(x)}{\cos(x)}\\
\end{array}
$
I will be glad to answer
any questions you may have.
