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! 
 A: 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 $\sin^2(\theta - \alpha) = \left(\dfrac 35\sin\theta - \dfrac45 \cos\theta\right)^2$
A: $$
\cos(\theta+\alpha) = \cos\theta\cos\alpha-\sin\theta\sin\alpha.
$$
We have $\tan\alpha=\dfrac43$ so if $\mathrm{opp}=4$ and $\mathrm{adj}=3$ then $\mathrm{hyp}=5$ by the Pythagorean theorem, so $\sin\alpha=\dfrac{\mathrm{opp}}{\mathrm{hyp}}= \dfrac45$ and $\cos\alpha=\dfrac35$.
Consequently $\cos^2(\theta+\alpha)=\left(\dfrac35\cos\theta+\dfrac45\sin\theta\right)^2$.
Do something similar in the denominator.  At some point you should probably multiply the top and bottom both by $5^2$ to clear out the fractions-within-fractions.
Next:
$$
1-\cos^2(\theta-\alpha) = \sin^2(\theta-\alpha),
$$
so we recall that
$$
\sin(\theta-\alpha) = \sin\theta\cos\alpha-\cos\theta\sin\alpha= \frac35\sin\theta - \frac45\cos\theta.
$$
Then we have
$$
\frac{\left(\dfrac35\cos\theta+\dfrac45\sin\theta\right)^2}{\left(\dfrac35\sin\theta - \dfrac45\cos\theta\right)^2} = \left(\frac{3\cos\theta+4\sin\theta}{3\sin\theta-4\cos\theta}\right)^2 = \left(\frac{3+4\tan\theta}{3\tan\theta-4}\right)^2.
$$
This is not constant as a function of $\theta$, but I surmise that what you mean is that one is to solve for $\theta$.  By a bit of simple algebra one goes from what you see above to $\tan\theta=\text{something}$, and from there to $\theta=\arctan\text{something}$.
A: Hint:
Draw a right angled triangle. You know the sides(How?)
A: $\tan\alpha=\frac{4}{3}$,then one solution of $\sin\alpha=\frac{4}{5},\cos\alpha=\frac{3}{5}$,the other is $\sin\alpha=-\frac{4}{5},\cos\alpha=-\frac{3}{5}$.
If your conclusion is right. The value of left hand does not depend on $\theta$, so we choose free, for easy calculation. I prefer $\theta=0$,the result is $\frac{9}{16}$. But when choosing $\theta=\pi/2$, the result  is $\frac{16}{9}$.
Could you check your problem again?
