1. ## Trig Identities

Not sure how to approach this problem:

2. $\displaystyle \frac{\sin{x}+\tan{x}}{\cos{x}+1}$

Multiply by $\displaystyle \frac{\frac{1}{\cos{x}}}{\frac{1}{\cos{x}}}$to get:

$\displaystyle \frac{\tan{x}+\frac{\tan{x}}{\cos{x}}}{1+\frac{1}{ \cos{x}}}$

Take $\displaystyle \tan{x}$ common factor, and we're done.

3. Originally Posted by casey_k
Not sure how to approach this problem:
Actually, with your working, you could've taken sin(x) common factor, then you can see a cancellation.

4. Hello, casey_k!

Another approach . . .

$\displaystyle \frac{\sin x + \tan x}{\cos x + 1} \;=\;\tan x$

We have: .$\displaystyle \frac{\sin x + \tan x}{\cos x + 1} \;=\;\frac{\sin x + \frac{\sin x}{\cos x}}{\cos x + 1} \;=\;\frac{\sin x(1 + \frac{1}{\cos x})}{\cos x + 1}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\;\;\frac{\cos x\cdot\sin x(1 + \frac{1}{\cos x})}{\cos x\cdot(\cos x + 1)} \;=\;\frac{\sin x(\cos x + 1)}{\cos x(\cos x + 1)} \;=\;\frac{\sin x}{\cos x} \;=\;\tan x$