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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 $1$?

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    $\begingroup$ For some basic information about writing mathematics, check out basic help on mathjax notation and mathjax tutorial and quick reference, $\endgroup$
    – Elliot Yu
    Nov 3, 2021 at 5:44
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    $\begingroup$ 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. $\endgroup$
    – Elliot Yu
    Nov 3, 2021 at 5:48
  • $\begingroup$ Actually, they are both correct, but not for the same values of $x$. Which also means they are both only partly correct. $\endgroup$ Nov 3, 2021 at 7:26

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

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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.

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  • $\begingroup$ Got it thankyou $\endgroup$
    – Neev Garg
    Nov 3, 2021 at 6:10
  • $\begingroup$ $\sqrt{1-\cos^2 x}=|\sin x|$, 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. $\endgroup$ Nov 3, 2021 at 7:33

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