# Homework Help: Trig Identity

1. Oct 8, 2005

### cscott

$$\cos \theta (\tan \theta + \cot \theta) = \csc \theta$$

2. Oct 8, 2005

### irony of truth

So, are you proving this identity?

Express your tangent and cotangent in terms of sine and cosine. Get their LCD... and your numerator becomes a well-known trigonometric identity..

Can you continue from here? :D

3. Oct 8, 2005

### cscott

The easy ones always get me :\

Thanks!

4. Oct 8, 2005

### cscott

I can't get this one either:

$$\frac{1 + \tan \theta}{1 - \tan \theta} = \frac{\cot \theta + 1}{\cot \theta - 1}$$

5. Oct 8, 2005

### mezarashi

For this one, you can either choose to replace tan x by 1/cot x or replace cot x by 1/tan x. Choose either and do some algebriac manipulations while leaving the other side alone.

6. Oct 8, 2005

### TD

Or, if that doesn't work for you, substitute tan by sin/cos and cot by cos/sin, then simplify the expressions

Try, if you get stuck, show us!

7. Oct 8, 2005

### cscott

I end up with

$$\frac{\cos^2 \theta + \sin \theta \cos \theta}{\cos^2 \theta - \sin \theta \cos \theta}$$

or

$$\frac{\cot^2 \theta + \cot \theta}{\cot^2 \theta - \cot \theta}$$

How do I continue?

8. Oct 8, 2005

### TD

How did you end up with that?

For the LHS:

$$\frac{{1 + \tan \theta }}{{1 - \tan \theta }} = \frac{{1 + \frac{{\sin \theta }}{{\cos \theta }}}}{{1 - \frac{{\sin \theta }}{{\cos \theta }}}} = \frac{{\frac{{\cos \theta + \sin \theta }}{{\cos \theta }}}}{{\frac{{\cos \theta - \sin \theta }}{{\cos \theta }}}} = \frac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}$$

Now try the RHS

9. Oct 8, 2005

### cscott

Silly me - I just multiplied out the numerator by the reciprocal of the denomenator instead of just canceling out the cosines. If you factor the top and bottom of my expression you end up with what your answer. If I do this using 1/cot = tan I end up with the RHS.

Don't I need to continue with the LHS until I get the right or vice versa?

10. Oct 8, 2005

### TD

Well now you have the LHS, the easiest would be trying to get the same starting with the RHS, which will go more or less the same

11. Oct 8, 2005

### cscott

Ah, I see. Thank you both of you.

12. Oct 8, 2005

No problem