SOLUTION: Prove the identity (sin(A-B)/sinAsinB)=cotB-cotA

Question 596372: Prove the identity
(sin(A-B)/sinAsinB)=cotB-cotA
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

There are a number of things to consider when trying to figure out these identities:

Match arguments -- Use argument-changing Trig properties (2x, 1/2x, A+B, A-B) to change arguments on the left to match those on the right
Match the number of terms -- Use algebra and/or Trig properties to get the same number of terms on the left side as there are on the right.
Match functions -- Use Trig properties to change functions on the left to match those on the right.
If none of the above look like they are going to work, try using Trig properties to change sec, csc, tan or cot on the left into sin and/or cos.

Unfortunately, I cannot give a recipe of what to do. You just have to know your properties and algebra well enough to see which of the above looks promising. (If nothing looks promising, change everything into sin's and/or cos's and look again.)

With your identity I see an argument of (A-B) on the left and no such argument on the left. So at some point we need to change that argument. That is where we will start. Using the sin(A-B) formula we get:

Now the arguments are all A's and B's.

The left side has one term while the right side has two. So we need to "split" the term on the left side in to two. Since it is a fraction that has a multiple term numerator we can "un-subtract":

Now we have the right number of terms on each side.

All that is left is to make those terms match. We should be able to see that we can reduce each of the fractions on the left side:

And we should recognize the fractions on the left as cot's"

And we are done!
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