# Another Trig Identity

1. Oct 9, 2005

### cscott

$$1 - \frac{\cos^2 \theta}{1 + \sin^2 \theta} = \sin \theta$$

I can only get to $$\frac{2\sin^2 \theta}{1 + \sin^2 \theta}$$ and I don't know if that's correct.

2. Oct 9, 2005

### TD

Perhaps you should check the problem again because it doesn't seem correct to me. Try $\theta = 3\pi /2$, the LHS will be 1 while the RHS is -1.

3. Oct 9, 2005

### Fermat

It's not an identity!

Let θ = -π/2

then lhs = 1 - cos²(-π/2)/(1 + sin²(-π/2))
lhs = 1 - 0/2
lhs = 1
=====

but rhs = sin(-π/2) = -sin(π/2) = -1
rhs = - 1
=======

Since lhs ≠ rhs, it can't be an identity.

4. Oct 9, 2005

### cscott

OMG, I've spent so much time trying to figure it out... I never thought of checking to make sure it was indeed an identity. I guess my teacher made a mistake. Thanks guys.

(Don't worry, I'll be back with another too soon :tongue2:)

5. Oct 9, 2005

### Fermat

I did a quick check to see if it was an identity. After all , you were having trouble !
It checked out ok for θ = 0 and θ = pi/2, so I did some work on it.
It was only later on that I plugged the expressions into graphmatica and found out they were actually two different functions !!

6. Oct 10, 2005

### cscott

$$\frac{1 - \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 - \sin \theta} = 2 \sec \theta$$

$$\frac{1 - \sin \theta}{1 + \sin \theta} = (\sec \theta - \tan \theta)^2$$

I can't seem to get anywhere with these...

7. Oct 10, 2005

### Fermat

For the 1st one, cross-multiply the two terms in the lhs and simplify.

8. Oct 10, 2005

### Fermat

For the 2nd one, multiply the lhs by (1 - sinθ) and simplify.

9. Oct 10, 2005

### cscott

I can only get to $2 \sin \theta + 2 \sin^2 \theta$ for the first after cross multiplying.

For the second, how can I just multiply by $(1 - \sin \theta)$? I thought you can only multiply by 1?

10. Oct 10, 2005

### arildno

$$\frac{1 - \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 - \sin \theta}=\frac{1 - \sin \theta}{\cos \theta}+\frac{1+\sin\theta}{1+\sin\theta}\frac{\cos\theta}{1-\sin\theta}=\frac{1 - \sin \theta}{\cos \theta} +\frac{1+\sin\theta}{\cos\theta}=\frac{2}{\cos\theta}=2sec\theta$$

$$\frac{1 - \sin \theta}{1 + \sin \theta} =\frac{1-\sin\theta}{1-\sin\theta}\frac{1-\sin\theta}{1+\sin\theta}=\frac{(1-\sin\theta)^{2}}{\cos^{2}\theta}=(sec\theta-tan\theta)^{2}$$

11. Oct 10, 2005

### cscott

I see, thanks.

12. Oct 10, 2005

### Fermat

Ah, sorry, that should have been, multiply above and below by (1 - sinθ)