# Trig Identity problem

1. Dec 1, 2006

### DethRose

Hey ive got an assignment on trig identities and cant figure this one out.

Prove the Identity:

tanx+cotx=2csc2x

I got to

tanx+cotx= 1/2 sin2x=1/4sinx2cosx

but when i get to the point where i have numbers in front of the sinx or cosx i dont know what to do.

Thanks for any help

2. Dec 1, 2006

### Kurdt

Staff Emeritus
try multiplying tan(x) + cot(x) by tan (x) and proceeding from there. I'll start you off,

$$tan(x) + cot(x) = \frac{sec^2(x)} {tan(x)}$$

3. Dec 1, 2006

Write the LHS in terms of $$\sin x$$ and $$\cos x$$. You should get:

$$\frac{1}{\sin x \cos x} = \frac{1}{\frac{1}{2}\sin 2x} = \frac{2}{\sin 2x} = 2\csc 2x$$

Last edited: Dec 1, 2006
4. Dec 2, 2006

### DethRose

im sorry but i still have no idea how to do this question

5. Dec 2, 2006

### bob1182006

have you tried changing everything to sin and cos ? doing that you will be able to get the right side to equal the left.

if you're still stuck just post what you've tried to do and someone could point out where you went wrong or give you some advice on what the next step would be

6. Dec 2, 2006

You know that $$\tan x = \frac{\sin x}{\cos x}$$ and $$\cot x = \frac{\cos x}{\sin x}$$

Therefore:

$$\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^{2} x + \cos^{2} x}{\sin x \cos x}$$.

$$\sin^{2} x + \cos^{2} x = 1$$ so we have $$\frac{1}{\sin x \cos x}$$.

Now we know the identity $$\sin 2x = 2\sin x \cos x$$. Thus $$\sin x \cos x = \frac{1}{2}\sin 2x$$.

So now we have $$\frac{1}{\frac{1}{2}\sin 2x}$$. Remembering that $$\frac{1}{\sin x} = \csc x$$ we can deduce that $$\frac{1}{\frac{1}{2}\sin 2x} = 2\csc 2x$$

Last edited: Dec 2, 2006
7. Dec 2, 2006

### DethRose

isnt it supposed to equal 2csc2x not 2cscx?

8. Dec 2, 2006

### Kurdt

Staff Emeritus
It does he probably made a typo. There are loads of ways to do it but basically all you have to do is find a page full of trig identities and fiddle about.

9. Dec 5, 2006

### Aviig

Is it incorrect to work both sides of the problem and meet in the middle, or do you need to get one side to completely equal another, because I can easily get both sides to be 1/sinxcosx.

10. Dec 5, 2006

You have to get one side completely equal to the other. Working both sides of the problem and "meeting in the middle" is not correct.

11. Dec 5, 2006

### Gib Z

And that is because in order to meet in the middle, your assuming the 2 sides are infact equal. If you mean change one side with identities, and change the other, and make them the same, then from there you can work backwards and get it from one side anyway.

12. Nov 3, 2008

### TrigMaster

OK this is a easy problem

L.S tanx+cotx
=Sinx/cosx+cosx/sinx

= sinx^2+cos^2/cosxsinx

=1/cosxsinx

R.S 2CSC2X
=2*1/2sinxcosx (The 2's here cancel out)
=1/sinxcosx

L.S=R.S

I tried to be as clear as possible, sorry but i'm not used to doing math on the computer :(