In order to prove
[tex]\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)[/tex]
Let's write both sides in terms of [tex]\sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x)[/tex] only.
Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write
[tex]\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)[/tex]
and
[tex]\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)[/tex]
So, their multiplication is
[tex]\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)[/tex]
So, the left hand side simplifies to
[tex]\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}[/tex]
Now, on with the right hand side. We have
[tex]1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}[/tex]
Now simply make this expression one fraction:
[tex]1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}[/tex]
And as you can see, the two sides are equal.
