Is there a way to find $\cos(a+b)$ and $\cos(a-b)$ in terms of $\tan a$ and $\tan b$?

Question

I know $\tan(a)$ and $\tan(b)$. Is there a way that I can find $\cos x$ and $\cos y$ in term of $\tan a$ and $\tan b$, where $x=a+b$ and $y=a-b$?

Thanks.

Answer 1

$\cos(a+b)=\cos a\cos b- \sin a \sin b$

Multiply by $\frac{\cos a\cos b}{\cos a\cos b}$

You'll get $$\cos(a+b)=\cos a\cos b- (\cos a\cos b)(\tan a \tan b)$$

Now substitute instead of $\cos a=\sqrt{\frac{1}{1+\tan^2 a}}$

And you'll get that $$\cos(a+b)=\sqrt{\frac{1}{(1+\tan^2 a)(1+\tan^2 b)}} (1-\tan a \tan b)$$

It is a bit complicated as you see, you can try and do some simplifications, and there might be another easier method to do that, but thats what came into my mind now.

Answer 2

Strictly speaking, no. As $\tan(x)=\tan(x+k\pi)$ for every integer $k$, you can only determine $\cos(a+b)$ and $\cos(a-b)$ up to a sign.