<s>[INST] Is there any other way to establish this trig identity? $\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $ I needed to verify this trig identity: $$\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)} $$
what I did is I worked on both sides individually:
LHS: $$\frac{\frac1{\cos(x)} + 1}{\frac {\sin(x)}{\cos (x)}} =\frac{1 + \cos(x)}{\sin(x)}$$
RHS: $$\frac{\sin(x)}{1 - \cos(x)}=\frac {\sin(x)(1-\cos^2(x))}{1 -\cos^2(x)}=\frac {\sin(x)(1-\cos^2(x))}{\sin^2(x)}= \frac{1 +\cos(x)}{\sin(x)} $$
QED.
I can't think of other way to verify this, like just working on one side.
Is there any?
 [/INST] $$\begin{align*} \frac{\sec x + 1}{\tan x} &= \frac{1 + \cos x}{\sin x} = \frac{(1 + \cos x)(1 - \cos x)}{\sin x (1 - \cos x)} = \frac{1 - \cos^2 x}{\sin x (1 - \cos x)} = \frac{\sin^2 x}{\sin x (1 - \cos x)} \\ &= \frac{\sin x}{1 - \cos x}. \end{align*}$$
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