If `tan theta tan phi=sqrt((a-b)/(a+b))`, prove that
`a-bcos 2theta)(a-b cos 2phi)` is independent of `theta` and `phi`.

Explanation

Let us put `tan theta=t_(1)` and `tan phi=t_(2)`. Then
`t_(1)^(2)t_(2)^(2)=(a-b)/(a+b)`
Also, `cos 2theta=(1-tan^(2)theta)/(1+tan^(2)theta)=(1-t_(1)^(2))/(1+t_(1)^(2))`
`cos 2phi=(1-tan^(2)phi)/(1+tan^(2)phi)=(1-t_(2)^(2))/(1+t_(2)^(2))`
Now, `a-b cos 2theta=a-b((1-t_(1)^(2)))/((1+t_(1)^(2)))`
`=((a-b)+(a+b)t_(1)^(2))/(1+t_(1)^(2))`
`=((a+b))/((1+t_(1)^(2)))[(a-b)/(a+b)+t_(1)^(2)]`
`=(a+b)/(1+t_(1)^(2))[t_(1)^(2)t_(2)^(2)+t_(1)^(2)]`
`=(1+b)/((1+t_(1)^(2)))(t_(2)^(2)+1)t_(1)^(2)`.
Similarly, `a-b cos 2phi=(a+b)/((1+t_(2)^(2)))t_(2)^(2)(t_(1)^(2)+1)`.
Hence `(a-b cos 2theta)(a-b cos 2phi)`
`=(a+b)^(2)t_(1)^(2)t_(2)^(2)=a^(2)-b^(2)`.
Which is independent if `theta` and `phi`.