Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ 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:


*

*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
$$\Longleftrightarrow\sin x(1+\cos x+\sin x)\equiv(1+\cos x)(1-\cos x+\sin x)$$
$$\Longleftrightarrow\sin x+\cos x\sin x+\sin^2 x\equiv1-\cos x+\sin x+\cos x-\cos^2 x+\sin x \cos x$$
$$\Longleftrightarrow\sin^2 x\equiv1-\cos^2 x$$
$$\Longleftrightarrow\cos^2 x +\sin^2 x\equiv1$$
$$\Longleftrightarrow \mathsf{true}$$

*Multiplying through by the "conjugate" of the denominator:
$${\rm\small LHS}\equiv\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} $$
$$\equiv\frac{1+\cos x + \sin x}{1 - (\cos x - \sin x)} ~~\cdot ~~\frac{1+(\cos x - \sin x)}{1 +(\cos x - \sin x)}$$
$$\equiv\frac{(1+\cos x + \sin x)(1+\cos x - \sin x)}{1 - (\cos x - \sin x)^2}$$
$$\equiv\frac{1+\cos x - \sin x+\cos x + \cos^2 x - \sin x \cos x+\sin x + \sin x \cos x - \sin^2 x}{1 - \cos^2 x - \sin^2 x + 2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- \sin^2 x}{2\sin x \cos x}$$
$$\equiv\frac{1+ 2\cos x + \cos^2 x- 1 + \cos^2 x}{2\sin x \cos x}$$
$$\equiv\frac{2\cos x (1+\cos x)}{2\cos x(\sin x)}$$
$$\equiv\frac{1+\cos x}{\sin x}$$
$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\equiv {\rm\small RHS}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$
Both methods of proof feel either inelegant or unnecessarily complicated. Is there a simpler more intuitive way to go about this? Thanks.

 A: For fun, I created a trigonograph:

$$\frac{1 + \cos\theta + \sin\theta}{1 + \sin\theta - \cos\theta} = \frac{1 + \cos\theta}{\sin\theta}$$
A: 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.
A: 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
$$
\frac{1+\cos x}{\sin x} = \frac{\sin x}{1-\cos x} =\frac{1+\cos x +\sin x}{1 -\cos x +\sin x}
$$
A: 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*}
A: alternatively, using abbreviations $c=\cos x$ and $s=\sin x$ we have
$$
s(1+c+s)=s(1+c) + s^2 = s(1+c) + 1-c^2=s(1+c)+(1-c)(1+c)=(1+s-c)(1+c)
$$
