# Quick trig problem

• Feb 14th 2010, 07:09 AM
elven06
Quick trig problem
Truth is, I haven't simplified trig problems in a while. I was wondering if you guys could help me out.

Here's the problem:

Write the trigonometric expression in terms of sine and cosine, and then simplify.

tan(x)/(sec(x)-cos(x))

• Feb 14th 2010, 07:16 AM
skeeter
Quote:

Originally Posted by elven06
Truth is, I haven't simplified trig problems in a while. I was wondering if you guys could help me out.

Here's the problem:

Write the trigonometric expression in terms of sine and cosine, and then simplify.

tan(x)/(sec(x)-cos(x))

note that $\tan{x} = \frac{\sin{x}}{\cos{x}}$

and $\sec{x} = \frac{1}{\cos{x}}$

• Feb 14th 2010, 07:26 AM
elven06
I tried, but I came out with:

sin(x)-(2/1+cos(x))

I'm not sure where I'm going wrong. Thanks for the quick reply.
• Feb 14th 2010, 07:38 AM
e^(i*pi)
Quote:

Originally Posted by elven06
I tried, but I came out with:

sin(x)-(2/1+cos(x))

I'm not sure where I'm going wrong. Thanks for the quick reply.

$\frac{\sin(x)}{\cos(x)} \times \frac{1}{\frac{1}{\cos(x)}-\cos(x)}$

$\sec(x)-\cos(x) = \frac{1-\cos^2(x)}{\cos(x)}$

$\therefore \: \: \frac{1}{\sec(x)-\cos(x)} = \frac{\cos(x)}{\sin^2(x)}$

$\frac{\sin(x)}{cos(x)} \times \frac{\cos(x)}{\sin^2(x)} = \frac{1}{\sin(x)} = \csc(x)$
• Feb 14th 2010, 08:41 AM
Soroban
Hello, elven06!

A slightly different approach . . .

Quote:

Write the trigonometric expression in terms of sine and cosine, and then simplify.

. . $\frac{\tan x}{\sec x-\cos x}$

We have: . $\frac{\dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos x} - \cos x}$

Multiply by $\frac{\cos x}{\cos x}\!:\quad \frac{\cos x\left(\dfrac{\sin x}{\cos x}\right) }{\cos x\left(\dfrac{1}{\cos x} - \cos x\right)} \;=\;\frac{\sin x}{1-\cos^2\!x} \;=\;\frac{\sin x}{\sin^2\!x} \;=\;\frac{1}{\sin x} \;=\;\csc x$

• Feb 14th 2010, 10:02 AM
elven06
Thanks guys, that was it. I had put 1/sinx, but it wanted csc...