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I know that $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ is correct. But having some hard time proofing it using trig relations. Some of the relations I used are $$ \sin x \cos x= \frac{1}{2} \sin(2 x)\\ \sin^2 x + \cos^2 x= 1\\ \sin^2 x = \frac{1}{2} - \frac{1}{2} \cos(2 x)\\ \cos(2 x) = 2 \cos^2(x)-1 $$ And others. I tried starting with multiplying numerator and denominator with either $\cos x$ or $\sin x$ and try to simplify things, but I seem to be going in circles.

Any hints how to proceed?

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

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  • $\begingroup$ I would recommend using the align environment to make the flow of your answer more clear, but I like the $\lozenge$ look of it too much. :) $\endgroup$ – Daniel W. Farlow Mar 11 '15 at 9:51
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    $\begingroup$ I left it as such because of the diamond actually :-) $\endgroup$ – TenaliRaman Mar 11 '15 at 10:01
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    $\begingroup$ I also like how the "middle fraction" dividing line is more bold too to better indicate the "symmetry." It's tempting to put $\frac{1+\sin x}{\cos x}$ on top of the $\lozenge$ structure to make it look more like a $\lozenge$ isn't it? Haha. If only! $\endgroup$ – Daniel W. Farlow Mar 11 '15 at 10:03
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    $\begingroup$ If only indeed! I tried to put something on top, but nothing I tried made much sense :D $\endgroup$ – TenaliRaman Mar 11 '15 at 10:36
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    $\begingroup$ Thanks for clearing that up! I really need to take time travel 101...I want to mess with some things and be born as Gauss. Maybe I can make that happen now! $\endgroup$ – Daniel W. Farlow Mar 11 '15 at 11:04
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You have to prove that $$ \frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}.\tag{1} $$ Note that $(1)$ may be expressed as follows by cross-multiplying: $$ \underbrace{(\cos x)(1+\cos x+\sin x)}_{\text{LHS}}=\underbrace{(1+\sin x)(1+\cos x-\sin x)}_{\text{RHS}}.\tag{2} $$ Now all you have to do is expand the LHS and RHS to see that they are equivlant:

We have \begin{align} \text{LHS} &= (\cos x)(1+\cos x+\sin x)\tag{definition}\\[0.5em] &= \cos x +\cos^2(x)+\cos x\sin x\tag{expand} \end{align} and \begin{align} \text{RHS} &= (1+\sin x)(1+\cos x-\sin x)\tag{definition}\\[0.5em] &= 1+\cos x-\sin x+\sin x+\sin x\cos x-\sin^2(x)\tag{expand}\\[0.5em] &= 1+\cos x+\sin x\cos x-\sin^2(x).\tag{simplify} \end{align} Now compare the LHS and RHS. We clearly have that $$ \cos^2(x)=1-\sin^2(x)\Longleftrightarrow \cos^2(x)+\sin^2(x)=1, $$ and we know this famous result to be true (or easily established otherwise). Hence, we have proved $(1)$.

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Rearrange and Cross multiply

it is required to prove that

$$\dfrac{\cos x +1 +\sin x}{1-\sin x +\cos x} = \dfrac{1+\sin x}{\cos x}$$ or

$$ \cos^2 x + \cos x (1 +\sin x ) = (1-\sin^2 x) + \cos x (1 +\sin x ) $$ or

$$ \cos x (1 +\sin x ) = \cos x (1 +\sin x ). $$

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  • $\begingroup$ That was cute. +1 $\endgroup$ – Daniel W. Farlow Mar 11 '15 at 9:57
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Hint:

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

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