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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
up vote 3 down vote accepted

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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