Verifying Trig Identities: csc(A-B)=secB

[gunnar14]

Homework Statement

Verify that each equation is an identity- directions
Problem- csc(A-B)=secB
---------------- <<< divide bar
sinA-cosAtanB

Homework Equations

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

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 canceled 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 I am stuck here... help please

 

[vela]

Homework Statement

Verify that each equation is an identity- directions
Problem- csc(A-B)=secB
---------------- <<< divide bar
sinA-cosAtanB

Homework Equations

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

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 canceled 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 I am 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.