Trig Identities

[gunnar14]

1. The problem statement, all variables and given/known data
Verify that each equation is an identity- directions
Problem- csc(A-B)=secB
---------------- <<< divide bar
sinA-cosAtanB
2. Relevant equations

well i tried to put in terms of sin cos and i've gotten stuck

3. The attempt at a solution

i started off by disributing i.e.>> cscA-cscB=secB/sinA-cosAtanB
next i changed tan to sin/cos and cancelled out cos so i was left with cscA-cscB=sec/sinA-sinB

changed csc to 1/sinA-1/sinB=sec/sinA-sinB

changed sec to 1/cos so i flipped and multiplied 1/sinA-sinB and got 1/sinA-1/sinB= 1/sincosA-sincosB and im stuck here..... help please

[vela]

1. The problem statement, all variables and given/known data
Verify that each equation is an identity- directions
Problem- csc(A-B)=secB
---------------- <<< divide bar
sinA-cosAtanB
2. Relevant equations

well i tried to put in terms of sin cos and i've gotten stuck

3. The attempt at a solution

i started off by disributing i.e.>> cscA-cscB=secB/sinA-cosAtanB
next i changed tan to sin/cos and cancelled out cos so i was left with cscA-cscB=sec/sinA-sinB

changed csc to 1/sinA-1/sinB=sec/sinA-sinB

changed sec to 1/cos so i flipped and multiplied 1/sinA-sinB and got 1/sinA-1/sinB= 1/sincosA-sincosB and im stuck here..... help please
Just to be clear, the problem you're trying to solve is to verify the identity

\csc(A-B)=\frac{\sec B}{\sin A-\cos A \tan B}

right? Your post is kind of hard to read.

Your very first step is wrong because \csc(A-B) \ne \csc A - \csc B. Try writing it in terms of sin(A-B) first and go from there.