# 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.

• Did you try multiplying everything by $1-\sin x$? – Julien Mar 7 '13 at 22:37
• You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Mar 7 '13 at 22:39
• I think some aspects of most of the answers below are more complicated than the really need to be. I've posted an answer that does it more simply. I don't take it all the way to the end, but the rest is routine. I am a worshipper of simplicity. – Michael Hardy Mar 8 '13 at 2:02

$$\frac{\cos x}{1-\sin x}-\tan x=\frac{\cos x}{1-\sin x}-\frac{\sin x}{\cos x}=$$

$$=\frac{\cos^2x-\sin x+\sin^2x}{\cos x(1-\sin x)}=\frac{1-\sin x}{\cos x (1-\sin x)}=\ldots$$

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.

Hint: $$\frac{A}{B}-\frac{C}{D}=\frac{AD-BC}{BD}$$

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 $$1=\sec^2x-\tan^2x=(\sec x-\tan x)(\sec x+\tan x)$$

$\frac{\cos x}{1-\sin x} - \tan x = \sec x$

$\displaystyle\frac{\cos x}{1-\sin x} - \frac{\sin x}{\cos x} = \sec x$

$$(\cos^2 x - \sin x (1 - \sin x)) / (1-\sin x) (\cos x) = \sec x$$

$$(\cos^2 x - \sin x + \sin^2 x) / (1-\sin x)\cos x = \sec x$$

$$\begin{array}{lll} (1 - \sin x) / ((1-\sin x)\cos x) &=& \sec x\\ 1/\cos x &=& \sec x\\ 1/\sec x &=& \sec x \end{array}$$

• You may want to check the FAQ section for directions on how to properly write mathematics in this site with LaTeX. They way you wrote it make it very difficult to understand. – DonAntonio Mar 7 '13 at 23:05
• The problem with this way of writing proofs of trigonometric identities (a way in which undergraduates write them all the time, and it's hard to talk them out of it) is that it's not made clear which direction the "If$\ldots$then$\ldots$" goes. This kind of argument is logically sound if you write at each step "This is true IF the line below is true." and then finally observe that the last line is true. A better way of writing the argument above is.......[continued below] – Michael Hardy Mar 8 '13 at 1:57
• . . . . . the following: $(\cos x/(1-\sin x))-\tan x$ $=(\cos x/(1-\sin x))-(\sin x/\cos x)$ $=(\cos^2 x-\sin x(1-\sin x))/((1-\sin x)\cos x)$ $=(\cos^2 x - \sin x + \sin^2x)/((1-\sin x)\cos x)$, and so on, going down the left column above until you get to $1/\cos x$, then go on: $=\sec x$, and if necessary (but it's not necessary in this case) continue with a string of equalities that go up the right column. – Michael Hardy Mar 8 '13 at 2:01
• I'm sorry about the formatting. I will read on how to make it look correctly here. Thanks. – nightcoder Mar 9 '13 at 20:11

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{#c00000}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \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}