<s>[INST] solve this problem of trigonometry. 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. 
 [/INST] 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:
$$
\cot B - \cot Y = \cot(A+Y) + \cot(A-B).
$$
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