# 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.

• Hint: substitute $U = A - B$ and $V = -(A+Y)$, then invert both sides of the equation. You'll see that you now have the same equation, but with $U, V$ instead of $B, Y$. – user3294068 Apr 24 '14 at 11:44
• Can u plz elaborate?And of course which equation are u talking of? – user142971 Apr 24 '14 at 12:01
• U have just replaced the terms with a symbol.How can this solve this problem? – user142971 Apr 24 '14 at 12:06

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:

$$\cot B - \cot Y = \cot(A+Y) + \cot(A-B).$$

• U r really great but how can i know when the problem demands replacement of a term with another like u did here to solve the problem? – user142971 Apr 25 '14 at 12:53