# Precalc - Trig Identities

I only need a hint as to where to go from here. My problem is this:

$$\dfrac{1+\tan(x)}{\sin(x)}-\sec(x)$$

Here's my work trying to solve the problem, up until I got stuck. Did I make a mistake somewhere or make it more complicated than it should have been?

$$\dfrac{(1+\tan(x))(\cos(x)}{\sin(x)\cos(x)}-\dfrac{\sin(x)}{\cos(x)\sin(x)}=\dfrac{(1+\tan(x))(\cos(x))-\sin(x)}{\cos(x)\sin(x)}=\dfrac{1+\tan(x)-\sin(x)}{{\sin(x)}}=\dfrac{1}{\sin(x)}+\dfrac{\dfrac{\sin(x)}{\cos(x)}}{\sin(x)}-\dfrac{\sin(x)}{\sin(x)}=\csc(x)+\sec(x)-1$$

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You neglected $\cos(x)$ in the second line. You didn't multiply $1 + \tan(x)$ by $\cos(x)$ –  NasuSama Mar 4 '14 at 3:33
Same technique as your other problems: write everything in terms of $\sin x$ and $\cos x$ (which means you should get rid of $\sec x$ and $\tan x$ and so on). Gather together fractions over a common denominator. Simplify. Use $\sin^x + \cos^2x = 1$. –  bubba Mar 4 '14 at 3:48

The problem is that you neglected $\cos(x)$ after combining the fractional expressions altogether. So from the end of the first line,
\begin{aligned} \dfrac{(1 + \tan(x))\cos(x) - \sin(x)}{\sin(x)\cos(x)} &= \dfrac{\cos(x) + \tan(x)\cos(x) - \sin(x)}{\sin(x)\cos(x)}\\ &= \dfrac{\cos(x) + \frac{\sin(x)}{\cos(x)}\cdot \cos(x) - \sin(x)}{\sin(x)\cos(x)}\\ &= \dfrac{\cos(x) + \sin(x) - \sin(x)}{\sin(x)\cos(x)}\\ &= \dfrac{\cos(x)}{\sin(x)\cos(x)}\\ &= \dfrac{1}{\sin(x)}\\ &= \csc(x) \end{aligned}
Let $c = \cos(x)$ and $s = \sin(x)$. Then the expression you gave is $$\frac{1 + s/c}{s} - \frac{1}{c} \\ = \frac{c+s}{cs} - \frac{1}{c} \\ = \frac{c+s}{cs} - \frac{s}{cs} \\ = \frac{c+s-s}{cs} \\ = \frac{c}{cs} \\ = \frac{1}{s} \\ = \csc(x)$$ This same technique will work with all your other similar problems. The only thing you need to know is that $c^2 + s^2 = 1$.