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