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

Solution 1:

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*}

Solution 2:

For fun, I created a trigonograph:

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

Solution 3:

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.

Solution 4:

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