It is given :
$$\sin(A-B)/\sin B = \sin(A + Y)/\sin (Y)$$
We have to prove $$\cot B - \cot Y = \cot(A + Y) + \cot(A - B).$$
Please help me solving this. I have tried to solve this by analyzing $$\cot B - \cot Y$$ From the given equation,we get $$\frac{\sin A \cos B - \cos A \sin B}{\sin B} = \frac{\sin A \cos Y + \cos A \sin Y}{\sin Y} $$ or $$\sin A \cot B - \cos A = \sin A \cot Y + \cos A$$ or $$\sin A(\cot B - \cot Y) = 2 \cos A $$ or $$\cot B - \cot Y = 2\cot A $$ but we have to prove $$\cot B - \cot Y = \cot(A + Y) + \cot(A - B).$$ I couldn't proceed further.
OK, you started with: $$ {\sin(A−B)\over \sin B}={\sin(A+Y)\over\sin Y} $$ Which led you to: $$ \cot B−\cot Y=2\cot A $$
Let $U = A - B$, which imples $B = A - U$.
Let $V = -(A+Y)$, which implies $Y = -(A+V)$.
Substitute these values into the first equation to get: $$ {\sin U \over \sin (A - U)} = {\sin (-V) \over \sin(-(A+V))} $$ Note that $\sin (-x) = - \sin (x)$, so the negative signs on the right half of the equation factor out, then cancel each other. Now invert the equations to yield:
$$ {\sin (A - U) \over \sin U} = {\sin (A + V) \over \sin V} $$
This equation is exactly the same as the first equation, but with U, V instead of B, Y. Thus, using the same logic, you can conclude:
$$ \cot U - \cot V = 2 \cot A $$ Substitute the definitions for $U$ and $V$: $$ \cot U - \cot V = \cot (A - B) - \cot (-(A+Y)) = \cot (A - B) + \cot (A + Y). $$
This is also equal to $2 \cot A$, so we can combine with the second equation above to yield:
$$ \cot B - \cot Y = \cot(A+Y) + \cot(A-B). $$