What is the value of $(\tan w)(\tan x)(\tan y)(\tan z)$?

Question

If $\frac{\cos(w-x)}{\cos(w+x)}+\frac{\cos(y+z)}{\cos(y-z)} = 0$

what is the value of $(\tan w)(\tan x)(\tan y)(\tan z)?$

I used cos formula.... But I could not figure it out...

Answer 1

simplifying your given term we get $$\cos(w-x)\cos(y-z)+\cos(y+z)\cos(w+z)$$ and we obtain $$\cos(x)\cos(y)\cos(z)\cos(w)+\sin(x)\sin(y)\sin(z)\sin(w)=0$$ Can you finish?

Answer 2

$$\dfrac{\cos(w-x)}{\cos(w+x)}=\dfrac{\cos(y+z)}{-\cos(y-z)}$$

Apply Componendo and Dividendo,

$$\dfrac{\cos(w-x)+\cos(w+x)}{\cos(w-x)-\cos(w+x)}=\dfrac{\cos(y+z)-\cos(y-z)}{\cos(y+z)+\cos(y-z)}$$

Use $\cos(A\pm B)$ formulae.