<s>[INST] 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.

 [/INST] For fun, I created a trigonograph:

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