Trigonometric Formula I am stuck with the simple expression
$$
\frac{\cos^2(\theta + \alpha)}{1 - \cos^2(\theta - \alpha)} = \text{const.}
$$
where $\theta$ is a variable and $\alpha$ is the number satisfying
$$
\alpha = \tan^{-1} (\frac{4}{3})\,.
$$
I cannot see it to be immediate, somehow I am missing a particular trigonometric identity. Or does this require a more detailed calculuation? Thanks for any hints, I'll fill in the details myself! 

Note that $1 - \cos^2(\theta - \alpha) = \sin^2(\theta - \alpha)$
That gives you:
$$\frac{\cos^2(\theta + \alpha)}{1 - \cos^2(\theta - \alpha)} = \text{const.} = \frac{\cos^2(\theta + \alpha)}{\sin^2(\theta - \alpha)}$$
For the numerator: $\cos(\theta+\alpha) = \cos\theta\cos\alpha-\sin\theta\sin\alpha.\tag{1}$
For the denominator: $\sin(\theta - \alpha) = \sin \theta \cos \alpha - \cos \theta \sin \alpha \tag{2}$
Note that since $\tan \alpha = \dfrac 43$, we have a $3:4:5$ triangle, using the Pythagorean Theorem, and noting that the leg opposite $\alpha$ must be of length $4$, and the leg adjacent to $\alpha$ is length $3$. This gives us a hypotenuse of length $5$.  Calculating $\cos \alpha = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac 35$. Likewise $\sin \alpha = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac 45.$
So $(1)$ becomes $\cos^2(\theta + \alpha) = \left(\dfrac 35 \cos\theta - \dfrac 45 \sin\theta\right)^2$.
And $(2)$ becomes $\sin^2(\theta - \alpha) = \left(\dfrac 35\sin\theta - \dfrac45 \cos\theta\right)^2$
