Find the value of $\big|\frac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\frac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\big|$ If $$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}=1,$$ where $\theta_1$ and $\theta_0$ do not differ by an odd multiple of $\pi$, then find the value of $$\left|\dfrac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\dfrac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\right|.$$
My attempt is as follows:
Attempt $1$:
$$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}$$
$$\dfrac{\sin\left(\theta_1+\theta_2\right)}{\cos\theta_2\sin\theta_2}=\dfrac{\sin\left(\theta_0+\theta_2\right)}{\cos\theta_2\sin\theta_2}$$
$$\sin\left(\theta_1+\theta_2\right)-\sin\left(\theta_0+\theta_2\right)=0$$
$$2\sin\dfrac{\left(\theta_1-\theta_0\right)}{2}\cos\dfrac{2\theta_2+\theta_1+\theta_0}{2}=0$$
either $\theta_1-\theta_0=2n\pi$  or  $2\theta_2+\theta_1+\theta_0=2n\pi$
Unfortunately it is given that $\theta_1-\theta_0$ is not equal to odd multiple of $\pi$ but we are getting $\theta_1-\theta_0$ as even multiple of $\pi$ so we cannot rule out one of the factor. Due to this reason I didn't find any way ahead.
Attempt $2$:
$$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=1$$
$$\sin\left(\theta_1+\theta_2\right)=\cos\theta_2\sin\theta_2$$
$$2\sin\left(\theta_1+\theta_2\right)=\sin2\theta\tag{1}$$
In the similary way
$$2\sin\left(\theta_0+\theta_1\right)=\sin2\theta\tag{2}$$
But by doing this we are tending towards result obtained in Attempt $1$:
Attempt $3$:
$$\left|\left(1-\dfrac{\sin\theta_0}{\sin\theta_2}\right)\left(1-\dfrac{\sin\theta_1}{\sin\theta_2}\right)+\dfrac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\right|$$
$$\left|1-\dfrac{\sin\theta_1}{\sin\theta_2}-\dfrac{\sin\theta_0}{\sin\theta_2}+\dfrac{2\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\right|$$
Also not getting anything from here, what to do?
 A: Hint:
Observe that $\theta_1,\theta_0$ are the roots of
$$\cos x\sin\theta_2+\sin x\cos\theta_2-\sin\theta_2\cos\theta_2=0$$
$$\iff\cos x\sin\theta_2=\sin\theta_2\cos\theta_2(\sin\theta_2-\sin x)$$
Now squaring both sides and replacing $\cos^2 x$ with $1-\sin^2x,$ we find
$$(1-\sin^2x)\sin^2\theta_2=\cos^2\theta_2(\sin\theta_2-\sin x)^2$$
$$\implies \sin^2x-\sin x\cdot 2\cos^2\theta_2\sin\theta_2-\sin^2\theta_2(1-\cos^2\theta_2)=0$$
So, $\sin\theta_1\sin\theta_0=-\dfrac{\sin^4\theta_2}1$
Similarly, $\cos\theta_1\cos\theta_0=?$
A: We have $$\sin\dfrac{\theta_0-\theta_1}2\cos\dfrac{\theta_0+\theta_1-2\theta_2}2=0$$
If $\sin\dfrac{\theta_0-\theta_1}2=0,$ this will make both equations identical
$$\implies\cos\dfrac{\theta_0+\theta_1-2\theta_2}2=0\implies\dfrac{\theta_0+\theta_1-2\theta_2}2=\dfrac{(2n+1)\pi}2$$  for some integer $n$
$\implies\theta_0+\theta_1=2n\pi+\pi-2\theta_2$
$$1=\left(\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}\right)\left(\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}\right) =\dfrac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\dfrac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}+\dfrac{\sin(\theta_0+\theta_1)}{\sin\theta_2\cos\theta_2}$$
Replace the value of $\theta_0+\theta_1$ in the last term
