# Simplify a trigonometric expression

I've been simplying a slew of trigonometric expressions and most of them fall out pretty clearly, but this one has been giving me fits:

$$\frac{(\sec x - \tan x)^{2} + 1}{\sec x\csc x - \tan x\csc x}$$

Here's my best attempt so far:

\begin{align*} &=\frac{\left(\frac{1}{\cos x}-\frac{\sin x}{\cos x} \right)^{2} + 1}{\frac{1}{\cos x\sin x}-\frac{1}{\cos x}}\\\\ &=\frac{\frac{1}{\cos^{2}x}-2\frac{\sin x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x} + 1}{\frac{\cos x-\cos x\sin x}{\cos^{2}x\sin x}} \end{align*}

But I can't help but think I've already gone wrong since I feel like I should start factoring things out again. Am I on the right track here?

Thanks for any suggestions.

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Try multiplying by $\frac{\cos^2 x}{\cos^2 x}$. – Brian M. Scott May 7 '12 at 3:25
You know $1+\tan^2 u=\sec^2 u$, yes? You can already use that in your numerator expression. – J. M. May 7 '12 at 3:28
Or multiply by $\frac {\cos^2x \sin x}{\cos^2x \sin x}$ to make it a single level fraction – Ross Millikan May 7 '12 at 3:33
As to your title, this is an expression not an equation that you are trying to simplify. – alex.jordan May 7 '12 at 3:35

There are certainly more efficient ways, as Ross and alex have already indicated, but if you want to continue on the path on which you've started, note that

\begin{align*} &\frac{\frac{1}{\cos^{2}x}-2\frac{\sin x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x} + 1}{\frac{\cos x-\cos x\sin x}{\cos^{2}x\sin x}}\cdot\frac{\cos^2x}{\cos^2x}\\\\ &\qquad=\frac{1-2\sin x+\sin^2x+\cos^2x}{\frac{\cos x(1-\sin x)}{\sin x}}\;. \end{align*}

Now use an obvious identity in the numerator, invert the denominator and multiply, and you're home free.

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Yes. Efficiency comes from experience :-) thanks. – Tom M May 7 '12 at 3:48

You have

\begin{align*} &=\frac{(\frac{1}{\cos x}-\frac{\sin x}{\cos x})^{2} + 1}{\frac{1}{\cos x\sin x}-\frac{1}{\cos x}}\\ &=\frac{\frac{1}{\cos^{2}x}-2\frac{\sin x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x} + 1}{\frac{\cos x-\cos x\sin x}{\cos^{2}x\sin x}} \end{align*}

Your next steps might be to note that on top, $\sin x/\cos x = \tan x$ and $1 + \tan^2 x= \sec^2 x$, leaving you with $2\dfrac{(1 - \sin x)}{\cos ^2 x}$

On the bottom, you have $\dfrac{\cos x}{\sin x}\dfrac{(1 - \sin x)}{\cos ^2 x}$

So cancel all that you can, and I think you'll be able to finish from here.

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We use $\sec^2 x-tan^2 x=1$, and replace the $1$ in the numerator by $\sec^2 x-\tan^2 x$. Then note the common factor $\sec x-\tan x$ appearing everywhere. Cancel it, noting the possible problem of cancelling $0$'s, or don't note it, because for reason unknown to me it is considered OK not to with trig identities.

After we cancel, we have $$\frac{(\sec x-\tan x)+(\sec x+\tan x)}{\csc x}.$$ Now we are essentially finished. The numerator is $2\sec x$, the denominator is $\csc x$. Replace them by $2/\cos x$ and $1/\sin x$ respectively, and we get $2\tan x$.

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I might start by factoring out the $\frac{1}{\csc x}$ (aka $\sin x$) and then multiplying by $\frac{\sec x+\tan x}{\sec x+\tan x}$, since $\sec^2x-\tan^2x =1$. This leaves

$$\sin(x)\left((\sec x-\tan x)^2 + 1\right)\left(\sec x+\tan x\right)$$

Distribute and repeat the identity...

$$\sin(x)\left(\sec x-\tan x + \sec x+\tan x\right)$$

And then we end up with $2\tan(x)$.

(Subtle note: The original expression was undefined for $x=n\pi$, since it involved $\csc(x)$. So really, this simplifies to $2\tan(x), x\neq n\pi$; its graph is like that of $2\tan(x)$, but with holes at every $x$-intercept.)

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