Trig Identities

(sin x * tan x) / (1 - cos x) - 1 = sec x
Show that left equals right.

$\begin{array}{rcll} \frac{ \sin{( x )} \cdot \tan{( x )} } { 1 - \cos{( x )} } - 1 & \overset{?}{=} & \sec {( x )} \qquad & | \qquad \sec {( x )} = \frac{1}{\cos {( x )}}\\\\ & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ \sin{( x )} \cdot \tan{( x )} } { 1 - \cos{( x )} } - 1 & \overset{?}{=} & \frac{1}{\cos {( x )}} \qquad & | \qquad \tan {( x )} = \frac{\sin {( x )} }{\cos {( x )}} \\\\ \frac{ \sin{( x )} \cdot \frac{\sin {( x )} }{\cos {( x )}} } { 1 - \cos{( x )} } - 1 & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ \sin^2{( x )} } { \cos {( x )}\cdot [ 1 - \cos{( x )} ]} - 1 & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ \sin^2{( x )} - \cos {( x )}\cdot [ 1 - \cos{( x )} ] } { \cos {( x )}\cdot [ 1 - \cos{( x )} ]} & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ \sin^2{( x )} - \cos {( x )} + \cos^2{( x )} } { \cos {( x )}\cdot [ 1 - \cos{( x )} ]} & \overset{?}{=} & \frac{1}{\cos {( x )}} \qquad & | \qquad \sin{( x )}^2 + \cos^2 {( x )} = 1 \\\\ \frac{ [ 1 - \cos {( x )} ] } { \cos {( x )}\cdot [ 1 - \cos{( x )} ]} & \overset{?}{=} & \frac{1}{\cos {( x )}} \\\\ \frac{ 1 } { \cos {( x )} } & = & \frac{1}{\cos {( x )}} \end{array}$