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?
 A: $$\begin{align*} \frac{\sec x + 1}{\tan x} &= \frac{1 + \cos x}{\sin x} = \frac{(1 + \cos x)(1 - \cos x)}{\sin x (1 - \cos x)} = \frac{1 - \cos^2 x}{\sin x (1 - \cos x)} = \frac{\sin^2 x}{\sin x (1 - \cos x)} \\ &= \frac{\sin x}{1 - \cos x}. \end{align*}$$
A: Working both sides of
$$\frac{\sec(x) + 1}{\tan(x)} = \frac{\sin(x)}{1 - \cos(x)}$$
First multiply both sides by $\sin(x)$ giving
$$\frac{\tan(x) + \sin(x)}{\tan(x)} = \frac{\sin^2(x)}{1 - \cos(x)}$$
Since $\sin^2(x) = 1-\cos^2(x)$ we immediately have
$$1 + \cos(x) = 1 + \cos(x)$$
A: Here's a bit of an off-beat approach, using the figure I call the Fundamental Trigonograph:

The right-hand and left-hand expressions involve sides from the similar $\sec$-$\tan$-$1$ and $1$-$\sin$-$\cos$ sub-triangles of the trigonograph.

Those triangles are in turn similar to the "$\operatorname{hyp}$-$\operatorname{opp}$-$\operatorname{adj}$" triangle, so that we can write
$$\frac{\sec\theta + 1}{\tan\theta} = \frac{\operatorname{hyp}\theta+\operatorname{adj}\theta}{\operatorname{opp}\theta}  \qquad\qquad
\frac{\sin\theta}{1-\cos\theta} = \frac{\operatorname{opp}\theta}{\operatorname{hyp}\theta-\operatorname{adj}\theta} \tag{1}$$
The equality of the fractions in $(1)$ follows from Pythagoras (where I'll suppress the "$\theta$"s to save typing and reduce clutter):
$$\operatorname{opp}^2 
= \operatorname{hyp}^2 - \operatorname{adj}^2
= \left(\operatorname{hyp} + \operatorname{adj}\right)\left(\operatorname{hyp} - \operatorname{adj}\right) 
\qquad\iff\qquad 
  \frac{\operatorname{hyp} + \operatorname{adj}}{\operatorname{opp}} 
= \frac{\operatorname{opp}}{\operatorname{hyp} - \operatorname{adj}} \tag{2}$$

Note that $(2)$ gives us not only the identity in question, but a whole slew of related results: Simply pick any two sub-triangles of the Fundamental Trigonograph and form the corresponding proportions. Compiling a complete list is left as an exercise to the reader, but here's one of the more-elaborate examples, using the largest triangle in the role of each sub-triangle:
$$\frac{\cot+\tan+\csc}{\sec} = \frac{\sec}{\cot+\tan-\csc} \tag{3}$$
Seeing the components of this identity amid the elements of the Fundamental Trigonograph ---so that we recognize this as just another instance of $(2)$--- is not necessarily easy. However, when it happens, it can save the hassle of the standard drill of converting everything to sines and cosines and whatnot.
A: \begin{align}
\frac{\sec(x) + 1}{\tan(x)}&=\frac{\cos(x) + 1}{\sin(x)}\\
&=\frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)}\\
&=\frac{2\cos(x/2)}{2\sin(x/2)}\\
&=\frac{2\cos(x/2)\sin(x/2)}{2\sin^2(x/2)}\\
&=\frac{\sin(x)}{1-\cos(x)}\\
\end{align}
A: Try doing the RHS in reverse. :-) 
A: Here's a less obvious approach.
$$\frac{\sec x + 1}{\tan x} = \frac{\sec x + 1}{\tan x}\cdot \frac{\sec x - 1}{\sec x - 1} = ?$$
