A Trigonometric Question If
$$\frac{\cos\left(x\right)}{\cos\left(\theta\right)}+\frac{\sin\left(x\right)}{\sin\left(\theta\right)}\:=1=\:\frac{\cos\left(y\right)}{\cos\left(\theta\right)}+\frac{\sin\left(y\right)}{\sin\left(\theta\right)}\:\:$$
Then what's the value of :
$$\frac{\cos\left(x\right)\:\cos\left(y\right)}{\cos^2\left(\theta\right)}+\frac{\sin\left(x\right)\:\sin\left(y\right)}{\sin^2\left(\theta\right)}$$ 
Options:
(a)2     (b)0     (c)1     (d)-1
**Solution given in my guide:** 
From Hypothesis:
$sin\left(x+θ\right)=sin\left(y+θ\right)$
Therefore
$2cos\left(\frac{x+y+2θ}{2}\right)sin\left(\frac{x-y}{2}\right)=0$
Now
$x-y\ne 2n\pi \:\:⇒\:sin\left(\frac{x-y}{2}\right)\ne \:0$  //How does this step work?
Hence
$\frac{x+y+2θ}{2}=\left(2n+1\right)\left(\frac{\pi }{2}\right)$
$x+y=\left(2n+1\right)\pi -2θ$
And proceeds to the final answer as -1.
 A: HINT:
Using Weierstrass substitution,
$$\frac{1-\tan^2\dfrac x2}{\left(1+\tan^2\dfrac x2\right)\cos\theta}+\frac{2\tan\dfrac x2}{\left(1+\tan^2\dfrac x2\right)\sin\theta}=1$$
On rearrangement we have a Quadratic Equation in $\tan\dfrac x2$
We shall reach the same equation for $y$
Hence $\tan\dfrac x2,\tan\dfrac y2$ are the roots of the  Quadratic Equation
A: $$\frac{\cos x}{\cos\theta}+\frac{\sin x}{\sin\theta}=1\iff\sin\theta\cos\theta=\cos x\sin\theta+\sin x\cos\theta$$
$$\iff\sin x\cos\theta=\sin\theta(\cos\theta-\cos x)$$
Squaring we get, $$(1-\cos^2x)\cos^2\theta=\sin^2\theta(\cos^2\theta+\cos^2x-2\cos\theta\cos x)$$
$$\iff\cos^2x-2\sin^2\theta\cos\theta\cos x+(\sin^2\theta-1)\cos^2\theta=0\ \ \ \ (1)$$
If we start with $\displaystyle\frac{\cos y}{\cos\theta}+\frac{\sin y}{\sin\theta}=1;$ we shall reach at the same equation.
So, the roots of $(1)$ are $\displaystyle\cos x,\cos y\implies\cos x\cos y=\frac{(\sin^2\theta-1)\cos^2\theta}1=-\cos^4\theta$
Similarly if we start with $$\frac{\cos x}{\cos\theta}+\frac{\sin x}{\sin\theta}=1\iff\cos\theta(\sin\theta-\sin x)=\cos x\sin\theta,$$
we shall reach at  $\displaystyle\sin x\sin y=-\sin^4\theta$
You should take it home from here
