How to show that $ \frac{\cos x}{1 - \sin x} - \tan x = \sec x$? Question is: verify the identity:
$$
\frac{\cos x}{1 - \sin x} - \tan x = \sec x.
$$
How do I show that the left side equals the right?
I changed $\tan x$ into $\sin x/\cos x$ but didn't get anywhere.
Please help.

$\frac{\cos x}{1-\sin x}  - \tan x = \sec x$
$\displaystyle\frac{\cos x}{1-\sin x} - \frac{\sin x}{\cos x} = \sec x$
$$(\cos^2 x - \sin x (1 - \sin x)) / (1-\sin x) (\cos x) = \sec x$$
$$(\cos^2 x - \sin x + \sin^2 x) / (1-\sin x)\cos x = \sec x$$
$$\begin{array}{lll}
(1 - \sin x) / ((1-\sin x)\cos x) &=& \sec x\\
1/\cos x  &=& \sec x\\
1/\sec x  &=& \sec x
\end{array}$$
