Trig Identities

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 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

Related Precalculus Mathematics Homework Help News on Phys.org
vela
Staff Emeritus
Homework Helper

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 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

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

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.