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.
 A: $$\frac{\cos x}{1-\sin x}-\tan x=\frac{\cos x}{1-\sin x}-\frac{\sin x}{\cos x}=$$
$$=\frac{\cos^2x-\sin x+\sin^2x}{\cos x(1-\sin x)}=\frac{1-\sin x}{\cos x (1-\sin x)}=\ldots$$
A: I don't normally like posting answers to elementary questions for which four answers are already here, but I think I can make this a bit simpler than the others.
$$
\frac{\cos x}{1-\sin x} = \frac{(\cos x)(1+\sin x)}{(1-\sin x)(1+\sin x)} = \frac{(\cos x)(1+\sin x)}{1-\sin^2 x}
$$
$$
= \frac{(\cos x)(1+\sin x)}{\cos^2 x} = \frac{1+\sin x}{\cos x}.
$$
The rest doesn't require anything but pushing on it till it's done.
A: Hint:
$$\frac{A}{B}-\frac{C}{D}=\frac{AD-BC}{BD}$$
A: $\frac{\cos x}{1-\sin x}  - \tan x = \sec x$
$\displaystyle\frac{\cos x}{1-\sin x} - \frac{\sin x}{\cos x} = \sec x$
$$(\cos^2 x - \sin x (1 - \sin x)) / (1-\sin x) (\cos x) = \sec x$$
$$(\cos^2 x - \sin x + \sin^2 x) / (1-\sin x)\cos x = \sec x$$
$$\begin{array}{lll}
(1 - \sin x) / ((1-\sin x)\cos x) &=& \sec x\\
1/\cos x  &=& \sec x\\
1/\sec x  &=& \sec x
\end{array}$$
A: hint
$$\frac{\cos x}{1-\sin x}-\tan x=\frac{\cos x/\cos x}{(1-\sin x)/\cos x}-\tan x=\frac{1}{\sec x-\tan x}-\tan x=\dots$$
also note that $$1=\sec^2x-\tan^2x=(\sec x-\tan x)(\sec x+\tan x)$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
{\cos\pars{x} \over 1 - \sin\pars{x}} - \tan\pars{x}}
=\bracks{{\cos^{2}\pars{x} \over 1 - \sin\pars{x}} - \sin\pars{x}}\sec\pars{x}
\\[5mm]&={\cos^{2}\pars{x} - \sin\pars{x} + \sin^{2}\pars{x}\over
          1 - \sin\pars{x}}\,\sec\pars{x}
={1 - \sin\pars{x}  \over 1 - \sin\pars{x}}\,\sec\pars{x}
=\color{#66f}{\large\sec\pars{x}}
\end{align}
