# Trigonometric Identity

1. Jul 7, 2008

### loisNominator

1. The problem statement, all variables and given/known data

I'm attempting to prove that

1 - sin^2 t /(1 + cos t) - cos^2/(1+tan t) = cos t sin t

2. The attempt at a solution

I've tried various approaches. The most promising has the LHS reduced to:

(sin t cos t (1 + cos t + sin t cos t))/((1 + cos t)(cos t + sin t)).

I've also shown numerically that the LHS resembles the RHS so I don't think there was a typo in the original problem.

Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 7, 2008

### dirk_mec1

Note that:

$$\frac{\sin^2(t)}{1+\cos(t)} = \frac{(1+\cos(t))(1-\cos(t))}{1+\cos(t)} = 1-\cos(t)$$

You think you can proceed from this point on?

3. Jul 8, 2008

### loisNominator

The first two terms simplify to cos t, but I'm still having the same issue wrt further simplification. One version has me stuck at:

cos t (cos t + sin t - cos^2 t)/(cos t + sin t)

I'd appreciate another hint. Thanks.

4. Jul 8, 2008

### Defennder

Is this the trigo identity you're trying to prove:

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

If so, then it doesn't appear to hold for $t=\frac{\pi}{4}$.

5. Jul 8, 2008

### loisNominator

So it appears. Let me go back to the person who set the original problem. Sorry!

6. Jul 8, 2008

### loisNominator

It turns out that it was a typo. It should read:
$$1 - \frac{\sin^2 t}{1 + \cot t} - \frac{\cos^2 t}{1+\tan t} = \sin t \cos t$$

Solution is straightforward.