# 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 $$1$$?

• For some basic information about writing mathematics, check out basic help on mathjax notation and mathjax tutorial and quick reference, Nov 3, 2021 at 5:44
• Also as to your actual question, I would encourage you to check your algebra for rationalizing the numerator. You should at first get a result that will still look different from rationalizing the denominator, but you can then try to prove that the two expressions are in fact equal. Nov 3, 2021 at 5:48
• Actually, they are both correct, but not for the same values of $x$. Which also means they are both only partly correct. Nov 3, 2021 at 7:26

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

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.

• Got it thankyou Nov 3, 2021 at 6:10
• $\sqrt{1-\cos^2 x}=|\sin x|$, and when $|\sin(x)|=-\sin(x)$, we have: $$\frac{1-\sin x }{\cos x}=\frac{1-\sin^2 x}{\cos x(1+\sin x)}=\frac{\cos}{1+\sin x}$$ So the OP is indeed almost right. Nov 3, 2021 at 7:33