# Trig Identities

• Nov 10th 2008, 07:03 AM
casey_k
Trig Identities
Not sure how to approach this problem:
• Nov 10th 2008, 07:13 AM
Chop Suey
$\frac{\sin{x}+\tan{x}}{\cos{x}+1}$

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

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

Take $\tan{x}$ common factor, and we're done.
• Nov 10th 2008, 07:19 AM
Chop Suey
Quote:

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.
• Nov 10th 2008, 08:05 AM
Soroban
Hello, casey_k!

Another approach . . .

Quote:

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

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