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It's required to prove that $$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$ I managed to go about out it two ways:

  1. Assume it holds: $$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$ $$\Longleftrightarrow\sin x(1+\cos x+\sin x)\equiv(1+\cos x)(1-\cos x+\sin x)$$ $$\Longleftrightarrow\sin x+\cos x\sin x+\sin^2 x\equiv1-\cos x+\sin x+\cos x-\cos^2 x+\sin x \cos x$$ $$\Longleftrightarrow\sin^2 x\equiv1-\cos^2 x$$ $$\Longleftrightarrow\cos^2 x +\sin^2 x\equiv1$$ $$\Longleftrightarrow true$$
  2. Multiplying through by the conjugate of the denominator: $$LHS\equiv\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} $$ $$\equiv\frac{1+\cos x + \sin x}{1 - (\cos x - \sin x)} ~~\cdot ~~\frac{1+(\cos x - \sin x)}{1 +(\cos x - \sin x)}$$ $$\equiv\frac{(1+\cos x + \sin x)(1+\cos x - \sin x)}{1 - (\cos x - \sin x)^2}$$ $$\equiv\frac{1+\cos x - \sin x+\cos x + \cos^2 x - \sin x \cos x+\sin x + \sin x \cos x - \sin^2 x}{1 - \cos^2 x - \sin^2 x + 2\sin x \cos x}$$ $$\equiv\frac{1+ 2\cos x + \cos^2 x- \sin^2 x}{2\sin x \cos x}$$ $$\equiv\frac{1+ 2\cos x + \cos^2 x- 1 + \cos^2 x}{2\sin x \cos x}$$ $$\equiv\frac{2\cos x (1+\cos x)}{2\cos x(\sin x)}$$ $$\equiv\frac{1+\cos x}{\sin x}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\equiv RHS~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$ Both methods of proof feel either inelegant or unnecessarily complicated. Is there a simpler more intuitive way to go about this? Thanks.
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    $\begingroup$ For (1), you are not assuming it holds, since you have all those $\iff$ steps. Not sure how it helps, but the right side is one of the half-angle formulae - $\cot(x/2)$. $\endgroup$ – Thomas Andrews Dec 26 '15 at 23:34
  • $\begingroup$ @ThomasAndrews That's true; but it's not a deductive $LHS \equiv \dots \equiv RHS$ type thing, which is what I was looking for. Interesting point about the half-angle formulae, thanks. $\endgroup$ – Luke Collins Dec 26 '15 at 23:37
  • $\begingroup$ Still, "assume" is wrong. You aren't assuming it, you are proving an equivalence and proving that each statement implies the other. Indeed, you don't even need the $\implies$ direction of that argument. $\endgroup$ – Thomas Andrews Dec 26 '15 at 23:42
  • $\begingroup$ @ThomasAndrews Right you are. $\endgroup$ – Luke Collins Dec 26 '15 at 23:46
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Since $1-\cos^2 x = \sin^2 x$, we have $f(x) := \dfrac{1+\cos x}{\sin x} = \dfrac{\sin x}{1-\cos x}$. Therefore,

\begin{align*}\dfrac{1+\cos x + \sin x}{1-\cos x + \sin x} &= \dfrac{f(x)\sin x + f(x)(1-\cos x)}{1-\cos x + \sin x} \\ &= \dfrac{f(x)[1-\cos x + \sin x]}{1-\cos x + \sin x} \\ &= f(x) \\ &= \dfrac{1+\cos x}{\sin x}.\end{align*}

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  • $\begingroup$ Well that's clever. Awesome :D $\endgroup$ – Luke Collins Dec 26 '15 at 23:39
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For fun, I created a trigonograph:

enter image description here

$$\frac{1 + \cos\theta + \sin\theta}{1 + \sin\theta - \cos\theta} = \frac{1 + \cos\theta}{\sin\theta}$$

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    $\begingroup$ That's beautiful! :) $\endgroup$ – Luke Collins Dec 27 '15 at 22:58
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Observe $$(1 - \cos x + \sin x)(1 + \cos x) = (1 - \cos^2 x) + (1 + \cos x)\sin x = \sin^2 x + (1 + \cos x)\sin x = (1 + \cos x + \sin x)\sin x,$$ from which the result immediately follows.

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  • $\begingroup$ Its simpler to start from the RHS $\endgroup$ – lab bhattacharjee Dec 27 '15 at 2:39
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if $$ \frac{a}{b}=\frac{c}{d}=k $$ then $$ \frac{a+c}{b+d} = \frac{kb+kd}{b+d} =k =\frac{a}{b} $$ since $$1-\cos^2 x =(1+\cos x)(1-\cos x) =\sin^2 x$$ we have $$ \frac{1+\cos x}{\sin x} = \frac{\sin x}{1-\cos x} =\frac{1+\cos x +\sin x}{1 -\cos x +\sin x} $$

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alternatively, using abbreviations $c=\cos x$ and $s=\sin x$ we have $$ s(1+c+s)=s(1+c) + s^2 = s(1+c) + 1-c^2=s(1+c)+(1-c)(1+c)=(1+s-c)(1+c) $$

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