We'll manipulate the left side of the equation:
[tex]\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\dfrac{\sin(x)\tan(x)}{1-\cos(x)}\cdot\dfrac{1+\cos(x)}{1+\cos(x)}\\\\
\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\dfrac{\sin(x)\tan(x)(1+\cos(x))}{(1-\cos(x))(1+\cos(x))}\\\\
\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\dfrac{\sin(x)\cdot\frac{\sin(x)}{\cos(x)}\cdot(1+\cos(x))}{1-\cos^2(x)}\\\\
\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\dfrac{\frac{\sin^2(x)}{\cos(x)}\cdot(1+\cos(x))}{\sin^2(x)}\\\\
\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\dfrac{\frac{1}{\cos(x)}(1+\cos(x))}{1}\\\\
[/tex]
[tex]\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\dfrac{(1+\cos(x))}{\cos(x)}\\\\
\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\dfrac{1}{\cos(x)}+\dfrac{\cos(x)}{\cos(x)}\\\\
\boxed{\dfrac{\sin(x)\tan(x)}{1-\cos(x)}=\sec(x)+1}~~\blacksquare[/tex]
